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Frank ICSE Solutions for Class 9 Chemistry – The Periodic Table

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Frank ICSE Solutions for Class 9 Chemistry – The Periodic Table

PAGE NO :132
Solution 1:
The present form of periodic table has eighteen vertical columns called ‘groups’. These groups are numbered from 1 to 18.

Solution 2:
Seven horizontal rows in periodic table are called periods.These are numbered as 1, 2, 3, 4, 5, 6 and 7 from top to bottom.

Solution 3:
Certain chemically similar elements are arranged into set of three elements are called triads. When these elements are arranged in order of their atomic mass, the atomic mass of the middle element is approximately average of the atomic masses of the other two.
For Example – Lithium, Sodium, Potassium

Solution 4:
Frank ICSE Solutions for Class 9 Chemistry - The Periodic Table 1

Solution 5:

  1. (a) When elements are arranged in increasing order of their atomic masses, the eighth element resembles the first in physical and chemical properties just like the eighth note on a musical scale resembles the first note.
  2. (b) It is based on the notes of the musical scale.
  3. (c) No, the law of octaves was not valid beyond calcium i.e. not valid for heavier elements having atomic no. greater than ’40’.
  4. (d) There are 6 elements in between A and B.

Solution 6:
Important conclusions of Newland’s law of octaves are-

  1. It was the first logical attempt to classify elements on the basis of atomic masses.
  2. Periodicity of elements was recognized for the first time.

Solution 7:
The physical and chemical properties of elements are the periodic functions of their atomic masses.

Solution 8:

  1. It helps to understand that the properties of elements are related with some fundamental unit of all elements.
  2. It simplifies and systemizes the study of the properties of various elements and their compounds.
  3. It helps to understand  the relationship between the different types of elements.
  4. It helps to understand the cause of gradual change in properties from one element to another.

Solution 9:
The number of elements in various periods are-
Period 1 – 2
Period 2 – 8
Period 3 – 8

Solution 10:
Hydrogen and Helium are the elements of period 1.

Solution 11:
At the end of the period 2 and period 3 the outer shell of the elements are fully filled i.e. they have complete octet.

Solution 12:
The element of group II (2) is likely to be metallic in character.

Solution 13:
Metallic character increases on moving down the group. Therefore, the element that is placed at the end of the group will have the greatest metallic character.

Solution 14:

  1. (a) Two alkali metals and their groups and period are-
    Lithium – First group and second period.
    Sodium – First group and third period.
  2. (b) Two alkaline earth metals with their groups and periods are-
    Beryllium – Second group and Second period
    Magnesium – Second group and Third period
  3. (c) Halogens belong to Group 17.
  4. (d) Elements of groups from 3 to 12 are called as transition elements.
  5. (e) Helium gas has duplet arrangement of electrons.
  6. (f) Germanium is a metalloid of group 14.
  7. (g) The bond formed between the atoms of an element of group 2 and group 16 is called as electrovalent or  ionic bond.
  8. (h) The bond formed between the atoms an element of group 14 and group 17 is called as covalent bond.
  9. (i) Lithium and Magnesium are the elements which are showing a diagonal relationship.
  10. (j) In Group 18 all elements are gases.

PAGE NO :133
Solution 15:

  1. (a) Newland, Newland law of Octaves
  2. (b) atomic masses
  3. (c) 18, 7
  4. (d) unstable
  5. (e) Atomic number, Moseley
  6. (f) similar outer electronic configuration at certain regular intervals
  7. (g) decreases
  8. (h) 2 electrons
  9. (i) different

Solution 16:

  1. (a) False
  2. (b) True
  3. (c) True
  4. (d) False
  5. (e) False
  6. (f) False
  7. (g) False

Solution 17:

  1. (a) A – 2, 1
    B – 2, 3
    C – 2, 8, 2
    D – 2, 8, 8
    E – 2, 8, 8, 2
  2. (b) B is placed right to A.
  3. (c) C and E belongs to same group.
  4. (d) The formula for sulphate of C is CSO4.
  5. (e) C has a smaller size atom.
  6. (f) Sodium is placed below A.
  7. (g) Helium is the first element of the group to which D belongs.
  8. (h) The formula for chloride of B is BCl3.
  9. (i) A2O is the formula of oxide of A.
  10. (j) Fluorine  is the non-metallic element which belongs to the same period as A.

Solution 18:

  1. (a) Properties of Germanium i.e. also called as eka-silicon were predicted on the basis of its position in Mendeleev’s periodic table.
  2. (b) Beryllium and Gold are the two elements whose atomic weights were corrected on the basis of their positions in Mendeleev’s periodic table.
  3. (c) There were 63 elements known at the time of Mendeleev’s classification of elements.

PAGE NO :134
Solution 19:
Merits of Mendeleev’s classification of elements are:-

  1. Systematic study of the element: Elements could now be studied as groups or families rather than individuals. The Mendeleev’s  periodic table simplified and systemized the study of the elements and their compounds.
  2. Prediction of new elements: While arranging the elements, in increasing order of atomic mass, Mendeleev left some blanks for the elements that were not discovered at that time. Though unknown, he was able to predict their properties more or less accurately.He named 3 of such elements as eka-boron, eka-aluminium, eka-silicon. He named them, as they were below boron,aluminium and silicon in the respective sub-groups.Eka-boron was later named as scandium, eka-aluminium as gallium and eka-silicon as germanium.
  3. Correction of atomic masses: The Mendeleev’s periodic table helped in correcting the atomic masses of elements based on their position in periodic table. Example-Atomic masses of Be was corrected from 13.5 to 9.

Solution 20:
Mendeleev left some gaps in his periodic table of elements for the elements that were not discovered at that time. Though they were unknown, he was able to predict their properties more or less accurately at that time.
Example – He named eka-boron, eka- aluminium and eka-silicon to some elements as they were just below boron, aluminium and silicon in the respective sub-groups. Eka-boron was later named as scandium, eka-aluminium as gallium and eka-silicon as germanium.

Solution 21:
The main characteristic of the last elements in the periodic table are-

  1. Their octet is attained.
  2. They are inert in nature.

The general name of such elements is noble gas or inert elements.

Solution 22:

  1. (a) The physical and chemical properties of elements are the periodic functions of their atomic numbers.
  2. (b) There are 7 periods and 18 groups  in the long form of the periodic table.
  3. (c) The number of elements in each period is-
    1 period-2
    2 period-8
    3 period-8
    4 period-18
    5 period-18
    6 period-32
    7 period-Incomplete

Solution 23:

  1. (a) X belongs to 17th group.
  2. (b) The period number of X is 3.
  3. (c) In atom of X there are 7 valence electrons.
  4. (d) Valency of X is 1.
  5. (e) It is non-metal.
  6. (f) The element right to it is Argon and the element left to it is Sulphur.
  7. (g) The atomic number of the elements above to it is 9 and below to it is 35 in the periodic table.

Solution 24:

  1. (a) The nature of bond will be ionic in the compound XY.
  2. (b)
    1. Y and Z will form covalent bond.
    2. X and Z will form ionic compound.

Solution 25:

  1. (a) There would be 2 electrons in the outermost shell of the element X.
  2. (b) Element X belong to second group of the periodic table.
  3. (c) When X reacts with chlorine, the compound formed of the formula XCl2.

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Frank ICSE Solutions for Class 9 Chemistry – Atomic Structure

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Frank ICSE Solutions for Class 9 Chemistry – Atomic Structure

PAGE NO :113
Solution 1:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 1

Solution 2:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 2

Solution 3:

  1. (a) Mass of an atom – Protons and neutrons
  2. (b) Size of an atom – Electrons

Solution 4:

Three fundamental particles of an atom are-

  1. Proton
  2. Electron
  3. Neutron

Solution 5:

  1. (a) An atom – An atom is the smallest particle of an element which may or may not be capable of independent existence.
  2. (b) An element – An element is usually defined as a pure substance that contains only one kind of particles. These particles may be atoms or molecules.

Solution 6:
Atomic number – The number of protons present in the nucleus of an atom is called the atomic number of the atom.
It is denoted by ‘Z’.

Solution 7:
The protons and neutrons are collectively called as nucleons.

Solution 8:
Isotopes are atoms of the same element, having the same atomic number, same chemical properties but different mass number, i.e., the atoms differ in the number of neutrons. The three isotopes of hydrogen atoms are-

  1. Protium
  2. Deuterium
  3. Tritium

Solution 9:
Electrons take part in a chemical reaction. Therefore, the chemical properties of an element depend upon the electronic configuration. Since, isotopes of an element have the same atomic number and hence same electronic configuration. So, they exhibit the same chemical properties.

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 3

Solution 11:
The outermost orbit of an element is called valence shell. The electrons present in the outermost orbit (valence shell) of an element are called valence electrons.

Solution 12:
Atoms may have incomplete octet. During the formation of a molecule, an atom of a particular element gains, loses or shares electrons until it acquires a stable configuration of eight electrons in its valence shell.

Solution 13:

  1. (a) Atomic number = Number of protons = 20
  2. (b) Mass number = Number of protons +Number of electrons = 20 + 20 = 40
  3. (c) Electronic configuration = 2, 8, 8, 2
  4. (d) Valency = 2

Solution 14:

  1. (a) A=1
    B=3
    C=2
    D=1
    E=1
  2. (b) E contains the greatest number of neutrons.
  3. (c) A contains the least number of electrons.
  4. (d) B contains equal number of electrons and neutrons.
  5. (e) A is a metal.
  6. (f) Fluorine is the most reactive of the non-metals.
  7. (g) Both are electronegative and non-metals.
  8. (h) D and E belongs to a particular family.
  9. (i) A3B will be the formula if atoms of A combines with atoms of B.
  10. (j) ‘B’ will combine in  a trivalent  element while ‘C’ will combine as a bivalent element.

PAGE NO :114
Solution 15:
“During the formation of the molecule, an atom of a particular element gains, loses electrons or shares electrons until it acquires a stable configuration of eight electrons in its valence shell” i.e. until it acquires octet.

Solution 16:
K can accommodate maximum of 2 electrons.
L can accommodate maximum of 8 electrons.
M can accommodate maximum of 18 electrons.

Solution 17:
Cathode rays are formed at the negative electrode of the discharge tube experiment.

Solution 18:

  1. In K maximum number of 2 electrons can be accommodated.
  2. In L maximum number of 8 electrons can be accommodated.
  3. In M maximum number of 18 electrons can be accommodated.
  4. In N maximum number of 32 electrons can be accommodated.

Solution 19:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 4

Solution 20:
Electronic configuration of magnesium is -2, 8, 2. Since, it has 2 electrons in its valence shell, so its valency is 2.

Solution 21:
Number of electrons in Sodium = 11
Number of protons in sodium = 11
Number of neutrons in sodium = 12
Number of nucleons in sodium = 23

Solution 22:
Inert elements are the elements which have completely filled valence shell. Since, they are already stable and do not need more electrons, they do not combine with other atoms. So, they exist as monoatoms in molecule.

Solution 23:

  1. (a) Hydrogen
  2. (b) K shell.
  3. (c) Magnesium
  4. (d) Isotopes
  5. (e) Helium has zero valency.

Solution 24:

  1. (a) The number of protons = 9
  2. (b) The number of neutrons = 19 – 9 = 10
  3. (c) The number of electrons = 9

Solution 25:
Atomic number is the number of protons of an atom which is unique to an atom but mass number is the total of number of protons and number of neutrons which may or may not be same to other atoms since there is probability of combination of number of protons and number of neutrons be same for two atoms.

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Frank ICSE Solutions for Class 9 Chemistry – Water

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Frank ICSE Solutions for Class 9 Chemistry – Water

PAGE NO :96
Solution 1:
The molecular formula of water is H2O, this shows that water is not an element but it is a compound made up of two elements hydrogen and oxygen  combined in a fixed ratio, i.e., 1:8 by mass. This also shows that the properties of water is totally different from those of hydrogen and oxygen as compounds have completely different properties than the elements of which it is made up of.

Solution 2:
Two characteristics properties responsible for making water a good solvent are:

  1. Water molecule can form hydrogen bonds with the molecules of other compounds.
  2. Water molecule is polar in nature.

Solution 3:
Frank ICSE Solutions for Class 9 Chemistry - Water 1

Solution 4:

  1. (a) Physical properties of water:
    1. Pure water is colourless, transparent, odourless and tasteless liquid.
    2. Boiling and Freezing points: At normal atmospheric pressure water boils at 100oC and freezes at 0oC.
    3. Pure water being a covalent compound is a very poor conductor of heat and electricity.On addition of electrolytes it becomes a good conductor of heat and electricity.
    4. Solvent properties:Water is a remarkable solvent dissolves many substances forming aqueous solutions because of its high dielectric constant.
    5. Anomalous behavior: On cooling water contracts in volume, as do other liquids, but at 4o C, it starts expanding, and continues to do till the temperature reaches 0oC,the point at which it converts to ice.
  2. (b)
    1. Boiling point of water: It increases with the addition of salt in water.
    2. Freezing point of water: It decreases with the addition of salt in water.
    3. Density of water: It increases with the addition of salt in water.

Solution 5:
Due to its high specific heat capacity, water is used in cooling systems for e.g., as coolant in motor car radiators.

Solution 6:
Frank ICSE Solutions for Class 9 Chemistry - Water 2

Solution 7:

  1. (a) Water reacts with metals oxides: corresponding hydroxides are formed which act as alkalis.
  2. (b) Water reacts with non metals oxides: corresponding acids are formed.

Solution 8:
Frank ICSE Solutions for Class 9 Chemistry - Water 3

Solution 9:
We can test the odourless, colourless liquid by:

  1. Measuring its boiling point, if comes out to be 100 oC then it confirms the liquid to be water.
  2. Measuring its melting point, if comes out to be 0 oC then it confirms the liquid to be water.

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Water 4

Solution 11:

  1. (a) Water being the universal solvent ,acts as an important mode of transport in plants,also water taken from natural source contains dissolved salts in them which is essential for the growth and development of plants
  2. (b) Water constitutes 3/4th of our human body and it is necessary for the survival of human beings and being the universal solvent helps in transportation and also regulates the body temperature.It also contains dissolved salts which supply essential minerals which are necessary for our body.

Solution 12:
Frank ICSE Solutions for Class 9 Chemistry - Water 5

Solution 13:
Frank ICSE Solutions for Class 9 Chemistry - Water 6

Solution 14:
Density of water is 0.997 gcm-3 at 4oC . The density of water decreases when the temperature increases above 4oC or decreases below 4oC.
At 40C, water has its maximum density and minimum volume. At any temperatur above or below 4 C, the density of water decreases. This property is called anomalous expansion of water. The significance of this unique property of water is that it enables marine life to exist in the colder regions of the world, because even when water freezes on the top, it is still liquid below the ice layer.

Solution 15:
At 40C, water has its maximum density and minimum volume. The property of anomalous expansion of water enables marine life to exist in the colder regions of the world because even when water freezes on the top, it is still liquid below the ice layer.

PAGE NO :97
Solution 16:
Frank ICSE Solutions for Class 9 Chemistry - Water 11

Solution 17:
Frank ICSE Solutions for Class 9 Chemistry - Water 7

Solution 18:

  1. (a) Liquids: alcohols, acids
  2. (b) Solids: Sugar, urea
  3. (c) Gases: Oxygen, carbon dioxide

Solution 19:
Rain water is the purest form of natural water.

Solution 20:

  1. (a) Solid in a liquid: Solubility decreases with rise in temperature in an exothermic process, for example, calcium sulphate(CaSO4), sodium sulphate(NaSO4), and in an endothermic reaction solubility increases with rise in temperature,for example potassium nitrte(KNO3) and sodium nitrate(NaNO3).
  2. (b) A gas in a liquid: Solubility decreases with rise in temperature.Thus, gases dissolves readily in cold water than in hot water.

Solution 21:
Frank ICSE Solutions for Class 9 Chemistry - Water 8

Solution 22:

  1. (a) Density of water is 0.997 gcm-3 at 4oC .The density of water decreases when the temperature decreases below 4oC. So ice has less density of 0.92 and is lighter  then that of water hence it floats on water.
  2. (b) A solution is said to be saturated if at a particular temperature if no more of the solute can be dissolved in it at that temperature, when the temperature is increased more solute particles can be dissolved in that saturated solution as the solubility of most of the substances generally increases with rise in temperature and thereby making it unsaturated.
  3. (c) Solubility of gases decreases with increase in temperature hence hot water contains less dissolved air than cold water.

Solution 23:

  1. (a) Solute: The substance which dissolves in a medium to produce a solution is called the solute.
  2. (b) Solvent: It is the medium in which the solute dissolves.
  3. (c) Solution: A homogeneous mixture that has a uniform composition throughout the volume of the mixture. It is the medium in which the solute dissolves.

Solution 24:

  1. (a) When we heat a saturated solution more solute can be dissolved in it, it as mostly solubility increases on raising the temperature and the solution then becomes an unsaturated solution.
  2. (b) When a hot saturated solution is cooled slowly and is kept undisturbed the excess salt does not separates out . Thus the solution becomes a supersaturated solution which contains more solute in it than it can hold at room temperature.
  3. (c) When the quantity of solvent increases,greater amount of solute can be dissolved.

Solution 25:
Frank ICSE Solutions for Class 9 Chemistry - Water 9

Solution 26:
Frank ICSE Solutions for Class 9 Chemistry - Water 10

Solution 27:
Solubility of a solid depends on:

  1. Size of particles: Smaller the size of particles of the solute, greater is the solubility.
  2. Contact between solute and the solvent: Increased contact between the solute and the solvent increases the solubility.
  3. Temperature: The variation of solubility with temperature depends on its nature i.e. whether the reaction is exothermic or endothermic.
    For exothermic reactions-The solubility decreases on increasing the temperature.
    For endothermic reaction-The solubility increases on increasing the temperature.

Solution 28:
Henry’s law states that:
At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

Solution 29:

  1. Z is least reactive so it will be found free in nature.
  2. X is more reactive so it will react with oxygen more readily and with ease.
  3. Z
  4.  Y is iron whose oxide is brown coloured Fe2O3.
  5. X
  6. The piece of Y i.e., iron is more reactive then copper so it will displace copper from copper nitrate and a brown layer of copper will be developed on the iron piece.

Solution 30:
Drop in, a crystal of the solute that is in the solution. If the crystal dissolves its unsaturated, if it stays the same size then its saturated and if it gets bigger than its supersaturated.

  1. This is because an unsaturated solution will be able to take in more solute, so it dissolves the crystal.
  2. A saturated solution will not take in more solute but will also not deposit any solute so the crystal stays the same size.
  3. A supersaturated solution wants to deposit its excess solute and become saturated, it just needs something to start it along.

The crystal starts it up and it will deposit its excess solute onto the crystal making the crystal bigger.

PAGE NO :98
Solution 31:

  1. (a) Increase in weight: when sodium chloride when exposed in air it gains moisture and its weight increases.
  2. (b) Decrease in weight: iron when exposed to air gets rusted due to the presence of moisture in the air it gets corroded, and hence its weight is reduced.
  3. (c) No effect on weight: when conc. Sulphuric acid when exposed to air  does not react and no change occurs in weight.

Solution 32:
Water pollution may be defined as the contamination of water by foreign substances which make it harmful for health of animals or plants or aquatic life, make it unfit for drinking and for domestic, industrial and agriculture use.

Solution 33:
Major pollutants of water are:
Ground water pollutants: Septic tanks, industry by products like pesticides, fertilizers, tanneries, mining wastes.
Surface water pollutants: No. of gases present in atmosphere like SO2, CO2, H2S, NOx, CO, etc., pollute surface water.
Lake water pollutant: organic wastes from hills, toxic effluents from urban areas, industrial effluents, dumping of huge amounts of sediments etc.,
River water pollutants: Industrial discharge, sewage discharges, detergents, discharges from drug, paper, and textile industries
Marine water pollutant: major pollutant is oil spills.

Solution 34:
Various sources of water pollution are:

  1. Natural process: Washing away of decomposed and animal wastes into main stream of water.
  2. Human activity: 
    1. (a)Discharge of Household detergents:  detergents used as a cleaning agent produce foam and pollute water. They do not undergo bio-degradation.
    2. (b) Discharge of industrial effluents: Industrial activities generate a variety of waste products which are generally discharged into water streams.The pollutants associated with the industrial effluents are  organic matter, inorganic dissolved salts, suspended solids. They inhibit oxidation or organic compounds; stabilize the colloidal impurities which do not aggregate to settle down.
    3. (c) Sewage:  Sewage is cloudy dilute aqueous solution containing minerals and organic matter.Sewage from homes and industries contains decomposable organic matter ,inorganic cations and anions,toxic metals etc.Pouring the drains and sewers in fresh water bodies causes water pollution.
      Water pollution due to sewage creates the following problems:

      1. Self purifying ability of water is lost and it becomes unfit for domestic purpose.
      2. Self regulatory capabilities of aquatic organism is retarded.
      3. Sewages produce pathogens which are diseases causing bacteria and result in water born gastro-intestinal diseases.

Solution 35:
Water treated for safe effluents involves sequential treatment:

  1. Primary treatment: The primary treatment involves physico-chemical processes to reduce settle able suspended solids of the wastes water and smoothened out individual effluent flow variations.It involves physic-chemical processes such as sedimentation,aeration,adsorption,oxidation etc.
  2. Secondary treatment: In this, the dissolved and colloidal organic matter present in waste is removed by biological processes involving bacteria and other micro organisms.
    These process may be aerobic or anaerobic.
    Aerobic treatment: Purification is carried out by aerobes in the presence of molecular oxygen.
    Anaerobic treatment: Purification of waste is achieved by anaerobes in complete absence of molecular oxygen.
  3. Tertiary treatment: It is the final treatment for polishing the effluents from secondary treatment. By this process suspended solids are removed, bacteria are removed, and organic and inorganic solids are removed.

Solution 36:
Two water born diseases are:

  1. Gastroenteritis
  2. Bacterial dysentery

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Frank ICSE Solutions for Class 9 Chemistry – Physical and chemical changes

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Frank ICSE Solutions for Class 9 Chemistry – Physical and chemical changes

PAGE NO :74
Solution 1:
A physical change is a temporary change in which no new substance is formed and the composition or identity of the substance is not altered although certain specific physical properties may be changed.

Solution 2:
A chemical change is a permanent change in which the original substance gives rise to one or more substances with different properties.

Solution 3:
The reactions in which heat is evolved are called exothermic reactions while the reactions in which heat is absorbed are called endothermic reactions.

Solution 4:

  1. False
  2. False
  3. False
  4. False
  5. True

Solution 5:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 1

Solution 6:
Possible conditions for a chemical change are-

  1. One or more new substance is formed during reaction.
  2. The change occurring during the reaction is permanent.
  3. The mass of the substance undergoing a chemical change is generally altered.
  4. Chemical change involves  making and breaking of bond.

Solution 7:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 2

Solution 8:

(a) Redox reaction – The reaction in which both oxidation and reduction takes place simultaneously is known as rtedox reaction. Oxidation is a reaction that involves the addition of oxygen or the removal of the hydrogen. Reduction is a reaction that involves the addition of hydrogen or the removal of oxygen.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 3
(b) Oxidation – Oxidation is a reaction that involves the addition of oxygen or the removal of the hydrogen. In electronic concept, it is defined as the process in which an atom, molecule or ion loses one or more electrons. This results in increase in the positive charge or decrease in negative charge on the resulting species.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 4
(c) Reduction – Reduction is a reaction that involves the addition of hydrogen or the removal of oxygen. In the electronic concept, it is defined as the process in which an atom, molecule or ion gains one or more electrons. This results in increase in the negative  charge or decrease in positive charge on the resulting species.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 5

Solution 9:

  1. Exothermic reaction
  2. Endothermic reaction

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 6

Solution 11:
The chemical reactions which occur with the absorption of light energy are called photochemical reactions.
Examples-
Decomposition of silver nitrate takes place in the presence of light.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 7

Solution 12:
As the burning substance combines with oxygen ,the total mass of the products should be greater than that of the burning substance. For example,when,magnesium is burnt,a new substance magnesium oxide is formed,whose weight is greater than that of the original magnesium.
Experiment – A crucible is weighed containing about 0.5 gm of magnesium.Now crucible is heated.When magnesium begins to burn,the lid is put back on the crucible and the lid is occasionally raised to allow air to enter and burn the magnesium such that no product is lost.When,all the magnesium has been burnt up, the crucible is allowed to cool and then on weighing it  we observe that there is gain in weight.

Solution 13:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 8

Solution 14:
Three conditions necessary for burning are-

  1. The substance to be burnt must be combustible.
  2. A supporter of combustion such as air or oxygen must be present.
  3. A combustible substance must be heated to its ignition temperature.

PAGE NO :75
Solution 15:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 9

Solution 16:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 10

Solution 17:
When oxidation occurs there is a loss of electrons but simultaneously there is a gain of electrons by other species which is called reduction. These both process occur simultaneously so we can say that both oxidation and reduction go hand in hand and such reactions are known as redox reaction.

Solution 18:

  1. Copper is oxidized to copper sulphate while sulphur in sulphuric acid is reduced to sulphur dioxide.
  2. Silver in silver oxide is reduced to silver while oxygen in hydrogen peroxide is oxidised to molecular oxygen.

Solution 19:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 11
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 12

Solution 20:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 13

Solution 21:
A candle is a stick of paraffin wax with cotton wick.As a candle burns wax melts and trickles down. It gets solidified shortly. This is physical change. Paraffin wax is a mixture of hydrocarbons. When wick catches fire, paraffin wax melts, evaporates and burns in air like any hydrocarbon to give carbon dioxide and water. This is a chemical change.

Solution 22:

  1. physical
  2. chemical
  3. chemical
  4. physical

Solution 23:
Two examples are:-

  1. Burning of wood-carbon get oxidized and oxygen gets reduced.
  2. Rusting-In it iron is oxidized.

Solution 24:

  1. Chromium(VI) .
  2. Hydrogen peroxide
  3. Barium carbonate.
  4. Silver nitrate.
  5. Manganese dioxide.

Solution 25:
Ignition temperature – Ignition temperature is the lowest temperature up to which temperature of a substance must be raised so that it catches fire.
A combustible substance must be heated to its ignition temperature for burning.

Solution 26:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 14

Solution 27:
On heating few crystals of iodine in a test tube, the grey crystals sublimes and dense violet fumes are seen. On cooling, the vapours again form the crystals. So, a physical change can be reversed.

Solution 28:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 20

Solution 29:
When water is freezed and evaporated, these both are physical changes because-

  1. The change is temporary and reversible.
  2. No new substance is formed and the chemical composition of the original substance remains the same.
  3. Mass of the substance remains unchanged
  4. The amount of energy required to bring about a physical change is generally equal to the amount of energy required to reverse the change. Hence,there is no net energy change involved.

Solution 30:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 15

Solution 31:
Air is necessary for burning. Incorrect amount of air in fuel combustion accounts for the largest losses in combustion system. If the fuel does not get enough air for combustion it will generate smoke and a potentially unhealthy mixture of gas products.

Solution 32:

  1. (a) Combustible substances -The substances that catch fire and burn easily. Ex-Wood, Charcoal, petrol, kerosene etc.
    Non-combustible substances-Substance which cannot burn in air or oxygen are called as non- combustible substances. Ex-Nitrogen gas, carbon dioxide etc.
  2. (b) Two substances other than oxygen that support combustion are-
    1. Hydrogen
    2. Nitrogen

Solution 33:

  1. (a)
    1. Burning of coal in air releases CO2 in air.
    2. Respiration releases carbon dioxide and water vapours.
  2. (b)
    1. Photosynthesis removes CO2 from the atmosphere. Plants take carbon dioxide from the atmosphere in the presence of sunlight and use it to synthesise glucose with the liberation of oxygen.
    2. Some man made chemical activities such as setting of mortar also use atmospheric carbon dioxide and helps in removing carbon dioxide.

Solution 34:
Nitrogen is inert in nature and does not support combustion while oxygen supports combustion.If proportions of nitrogen and oxygen in the air were reversed then the rate of combustion of substances will increase.

Solution 35:
Heating of sulphur – If some powdered sulphur is heated gently in a glass test tube, it melts to a pale yellow liquid. Flame is removed to stop heating, it is quickly changed back to solid sulphur.

PAGE NO :76
Solution 36:
Activity series – The arrangement of the metals in the decreasing order of their chemical reactivity is called the activity series.
In displacement reactions, a more reactive element (metal or non-metal) displaces a lesser reactive element from its compound. With the help of the activity series, it is possible to predict which metals will displace other metals from their solutions.

Solution 37:
Balance of oxygen and carbon dioxide is maintained in nature because there is a natural oxygen cycle and a natural carbon cycle operating all the time by which the desired proportions of the two gases in the air are maintained. This is also known as carbon dioxide-oxygen cycle.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 16

Solution 38:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 17

Solution 39:
Carbon dioxide from the atmosphere enters the plant through photosynthesis, where carbohydrates are produced. From green plants, the carbon in the form of carbohydrates, etc. enter the animal and human bodies. The atmospheric carbon dioxide gets dissolved in oceans by diffusion. Marine algae and photosynthetic bacteria obtain carbon dioxide from water.
Carbon dioxide returns to the atmosphere by respiration, combustion of fossil fuels like coal, wood, petroleum etc., weathering of rocks, volcanic eruptions etc.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 18

Solution 40:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 18

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Frank ICSE Solutions for Class 9 Chemistry – The Language of Chemistry

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Frank ICSE Solutions for Class 9 Chemistry – The Language of Chemistry

PAGE NO :56
Solution 1:
Symbol – It is the short form or abbreviation used for the name of an element. It represents one atom of that element.
Formula – Formula of a compound represents the composition of a molecule of the substance in terms of the symbols of the elements present in the molecule.

Solution 2:

  1. CaCO3
  2. MgSO4
  3. Fe2 (SO4)3
  4. CaHCO3
  5. CuI
  6. K2Cr2O7
  7. KMnO4
  8. Na2SO4
  9. Mg (NO3)2
  10. Ca3 (PO4)2

Solution 3:

  1. Valency – The combining capacity of an element is called its valency.
  2. Helium < Sodium < Magnesium < Carbon < Phosphorous

Solution 4:
Law of conservation of matter governs a completely balanced equation. It states that “matter can neither be created nor destroyed.”

Solution 5:
A symbol signifies one atom of that element.

Solution 6:
Latin names of the following compounds are-
Iron                      –   Ferrum
Tin                        –   Stannum
Lead                     –   Plumbum
Sodium                –   Natrium
Potassium           –   Kalium
Mercury              –   Hydragyrum

Solution 7:
The equation in which the total number of atoms of each element in the reactants, on the left side of the equation is same as the number of atoms in the products formed, on the right side of the equation is called as balanced chemical equation.

Solution 8:
Frank ICSE Solutions for Class 9 Chemistry - The Language of Chemistry 1

Solution 9:
A chemical equation gives information about-

  1. What substances enter into a given reaction (reactants) and what products are formed as a result of the reaction.
  2. The  quantities of the reactants and the product formed.
  3. The  optimum conditions of temperature and pressure.

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - The Language of Chemistry 2

Solution 11:
H2 means hydrogen, O4 means oxygen and S means sulphur  in the formula of H2SO4.

Solution 12:

  1. The highest valency of the element Z is six.
  2. The formula of the fluoride of Z will be ZF6.

Solution 13:
The three valencies of element are-

  1. Two-Since the element X combines with two hydrogen to form H2X and two atoms of X combines with one carbon to form CX2.
  2. Four-Since the element X combines with two oxygen to form XO2.
  3. Six-Since the element X combines with three oxygen to form XO3.

Solution 14:
Variable valency – Some elements are capable of showing more than one valency in their compounds called variable valency.
Some elements show variable valency i.e. more than one valency since these elements have more than one common valency state.

Solution 15:
Chemical formula – It represents the composition of a molecule of the substance in terms of the symbols of the elements present in the molecule. The rule for writing the formula is criss-cross method.

  1. The positive and negative radicals are represented by their symbols and written side by side with the correct valency written below each.
  2. The valencies are divided by their highest common factor if any to get the simplest ratio.
  3. These numbers are then interchanged and written as subscripts.

Solution 16:
Frank ICSE Solutions for Class 9 Chemistry - The Language of Chemistry 3

Solution 17:

  1. Sodium hydrogencarbonate
  2. Sodium hexacyanoferrate(III)
  3. Manganese(II) borate
  4. Calcium phosphate
  5. Potassium manganate(VI)

Solution 18:
Co stands for cobalt which is an element while CO stands for carbon monoxide which is a compound.

Solution 19:
Radical – A radical is an atom or a group of atoms of same or different elements that behaves in the manner of positive or negative ion. Radicals have their own combining power(valency) and chemical formulae.
Examples-
Monovalent  radicals -H–,OH–,Cl–,NO3–,H+,Na+,K+,NH4+
Trivalent radicals-PO43- ,Fe(CN)63-,AsO3 3-,N3-,Fe3+,Al3+,Bi3+,Au3+

Solution 20:
Frank ICSE Solutions for Class 9 Chemistry - The Language of Chemistry 4

Solution 21:

  1. Anion – Negatively charged radicals are termed as anions.
  2. Cation – Positively charged radicals are termed as cations.

Solution 22:
Disadvantages associated with hit and trial method of balancing of equations-

  1. It is tedious and takes a long time.
  2. The method is rather difficult for balancing such equations which contain the same element being repeated in a number of compounds.
  3. It does not give any information regarding the mechanism of the reaction.

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Frank ICSE Solutions for Class 9 Chemistry – Elements, Compounds and Mixtures

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Frank ICSE Solutions for Class 9 Chemistry – Elements, Compounds and Mixtures

PAGE NO :41
Solution 1:

  1. Elements: An element is a pure substance which can neither be broken down into simpler substances nor formed from two or more simpler substances by any known physical or chemical process. It is made of only one kind of atoms. It can be divided into four main categories
    1. Metals-Iron, magnesium
    2. Non-metals-Hydrogen, oxygen
    3. Metalloids-Arsenic, antimony
    4. Noble gas-Helium, neon
  2. Compound: A compound is a pure substance that is composed of two or more elements chemically combined in definite proportion by mass.
    The physical and chemical properties of a compound are different from those of its constituent elements.Hydrogen gas is combustible and oxygen is supporter of combustion , their  compound water  which is liquid is neither combustible nor a supporter of combustion.
  3. Mixture: Mixture is a physical combination of two or more substances, whether elements or compounds, which are mixed in any proportion by mass and retain their original properties even after mixing.
    Homogeneous mixture: They have same composition  and the same properties throughout their entire mass. Example- Salt solution, alloys etc.
    Heterogeneous mixture: They have different composition and different properties in different parts of their mass. Example-Mixture of sand and salt, mixture of iron fillings and sulphur etc.

Solution 2:

  1. Oxygen
  2. Carbon, hydrogen, Oxygen
  3. Mercury, Bromine
    1. Helium
    2. Oxygen
  5. Gallium,caesium
  6. Two noble gases are-
  7. Helium
  8. Argon

PAGE NO :42
Solution 3:
Air is a mixture because-

  1. The composition of air is not fixed i.e. the components may be present in any proportion by mass.
  2. Components of air i.e. nitrogen, oxygen etc. do not react with each other.

Solution 4:
Elements – Lead, Mercury, Sodium
Mixtures – Air, petrol, ink, gunpowder
Compounds – Common salt, alcohol, sand

Solution 5:
Pure substance – A pure substance is one which is made up of only one kind of particles. These particles may be atoms or molecules.
Example-Sulphur,  water.
Impure substance – They are mixtures of two or more chemically different substances mixed in indefinite proportions. The constituent substances retain their properties in the mixture.
Example-Mixture of salt and sand, gunpowder

Solution 6:
Mercury is the metal which is liquid at room temperature and bromine is the non-metal which is liquid at room temperature.

Solution 7:
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 1

Solution 8:
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 2

Solution 9:
Two reasons for believing that copper is a metal and sulphur is a non-metal are:-

  1. Copper is malleable and ductile while sulphur is neither malleable nor ductile.
  2. Copper is a good conductor of heat while sulphur is not good conductor of heat.

Solution 10:
Metalloids – The elements which possess properties intermediate between those of the metals and non-metals are called as metalloids. They react with both acids and alkali’s to form salts.
Ex – Arsenic, antimony

Solution 11:
Graphite is a non-metal which is a good conductor of electricity.

Solution 12:
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 3

Solution 13:
Mixture is the general name of the materials which contain atleast two pure substances and show the properties of their constituents.

Solution 14:

  1. Sodium
  2. Bromine
  3. Arsenic
  4. Radon
  5. Mercury
  6. Oxalic acid
  7. Carbon dioxide

Solution 15:
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 4

Solution 16:

  1. simpler substances
  2. atomic
  3. same
  4. mixture of salt and water
  5. two

Solution 17:
Names of two other mixtures which contain elements only are-

  1. Bronze
  2. Duralumin

Solution 18:

  1. A Molecule – The smallest particle of a substance that retains the chemical and physical properties of the substance and is composed of two or more atoms.
  2. Atomicity – Atomicity of an element is defined as the number of atoms present in one molecule of that element.

Solution 19:
Since, the constituents of a mixtures may be present in varying proportions so it cannot be expressed by a fixed chemical formula.

Solution 20:

  1. Air
  2. Cement
  3. Milk Sugar solution

Solution 21:
If a mixture of powdered iron and sulphur is heated in a test tube, a black shiny compound iron(II) sulphide (FeS)  is formed.

Solution 22:

  1. Tungsten, Mercury
  2. Graphite, Iodine

PAGE NO :43
Solution 23:
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 5

Solution 24:

  1. Chromatography – The chromatography is a technique of separating pure substances from the mixture.
    Advantages of chromatography –

    1. It requires a very small amount of the substance or sample.
    2. The components retain their individuality during the process.
    3. Chromatography finds application in easy separation of substances with similar physical and chemical properties.
  2. Filtration – It is a separation technique for separating a mixture in which one component should be solid and insoluble in the other liquid component.
    Example- Barium sulphate  in water.
  3. Fractional distillation – It is a technique used to separate  two liquids which dissolve in one another.The separation relies on the differences in boiling points of the two liquids.
    No, mixture of chloroform and water cannot be separated by this method.
  4. Centrifugation – It is a method for separating the suspended particles of a substance from a liquid in which the mixture is rotated at a high speed in centrifuge machine.
    Application – The clay particles in water (which are very fine) can be separated by centrifugation.

Solution 25:

  1. The vapour state which is obtained by heating solid without passing through liquid state is called sublimate.
  2. A liquid condensed from vapour in distillation is called distillate.
  3. The liquid produced after filtering a suspension of a solid in a liquid is called filtrate.
  4. Supernatant liquid is the upper layer of fluid found after a mixture has been centrifuged.
  5. If there is a heterogeneous mixture containing an insoluble solid in a liquid, then the solid substance that settle down is called sediment.

Solution 26:
We use fractional distillation to separate alcohol from a mixture of alcohol and water since the difference in boiling point between alcohol and mixture is very less.

Solution 27:

  1. We obtain pure water from sea water by distillation.
  2. A sample of pure iodine and sodium chloride is obtained by sublimation.

Solution 28:
The separation of the mixture depends upon-

  1. Size of the constituents
  2. Magnetic properties of constituents
  3. Mass of the constituents
  4. Solubility of the constituents
  5. Miscibilities of the constituents
  6. Boiling point of the constituents
  7. Diffusion rate of the constituents

Solution 29:
This is a separation technique of solid-solid mixture. This method involves the use of a solvent in which only one of the solid present in the mixture dissolves. Undissolved solid is removed by filtration. Mixture of ammonium chloride and silver chloride is separated by this method.

Solution 30:

  1. By distillation and fractional distillation we separate the mixture of two liquids.
  2. Yes, mixture of chloroform (B.P.= 61 oC) and carbon tetrachloride (B.P.=77 oC) be satisfactorily separated by the process of fractional distillation which is used for separating the various fractions of petroleum.
    For this purpose we will make two fractionating columns in the apparatus.

Solution 31:

  1. Solid-solid mixtures
    1. Magnetic separation method-Separation of iron ore from impurities
    2. Gravity separation-Mixture of saw dust and sand
    3. Solvent extraction-Mixture of sulphur and sand
  2. Solid- liquid mixtures
    1. Evaporation-Water and sodium chloride
    2. Distillation-Iodine in chloroform
    3. Filtration-Barium sulphate in water
  3. Liquid-liquid mixtures
    1. By separating funnel-Oil and water mixture
    2. Distillation-Acetone and water
    3. Fractional distillation-Ethyl alcohol and water

Solution 32:
The chromatography is the technique of separating pure substances from the mixtures. The chromatographic techniques was first employed by a Russian scientist Michael Tswett in 1903 for the separation of coloured substance from the mixture.
Principle of chromatography: The principle of chromatography is based on the difference in the extent of interaction (absorption) of various substances with a stationary phase and a mobile phase. A substance which interacts strongly with the mobile phase goes ahead of the other substance which interacts strongly with the stationary phase.

Solution 33:
Ink generally contain more than one dye. This mixture of ink is used as moving phase. Different constituents of ink move at different speed. The solvent rises up the filter paper over the spot and carries the different coloured components of ink to different heights on the filter paper. Each spot thus obtained at a particular height on the filter paper contains a particular constituent of the ink. Thus, the components of the ink are separated.

Solution 34:
By the help of fractional distillation we separate the components of liquid air.

Solution 35:

  1. Increase in weight – Sulphuric acid and iron
  2. Decrease in weight – Sodium carbonate crystals
  3. No change in weight – Sodium chloride

Solution 36:
By filtration, we will separate a mixture of chalk powder and water.

Solution 37:
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 6

Solution 38:
Fractionating column avoid the collection of distillate and re-distillation of distillate several times during fractional distillation.

Solution 39:
Two pair of liquids which can be separated by using a separating funnel-

  1. Oil and water
  2. Chloroform and water

Solution 40:
At first, with the help of magnet, iron nails will separate. Then, by sublimation camphor will separate from common salt.

Solution 41:
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 7
Frank ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures 8

PAGE NO :44
Solution 42:

  1. When a magnet is moved over ‘X’, iron fillings are pulled away and stick to the magnet. When a magnet is moved over ‘Y’, it remained  unaffected.
  2. When’X’ is treated with carbon disulphide, sulphur dissolves but not iron. While, when’Y’ is treated with carbon disulphide, iron sulphide does not dissolve  but sinks to the bottom of the test tube.
  3. When ‘X’ is treated with dilute HCl, a colourless, odourless gas hydrogen is evolved which burns with a blue flame and is extinguished with a pop sound. While, when ‘Y’ is treated with dilute HCl, a colourless gas with the smell of rotten eggs is evolved which is H2S.
    There is difference in the behavior of ‘X’ and ‘Y’ because ‘X’ is a mixture while ‘Y’ is a compound. The component of a mixture do not react chemically, so retain their identity in the mixture while the components of compound react chemically, so do not retain their identity in the compound.

Solution 43:
Centrifugation is used in milk dairies to separate cream from milk dairies.

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Frank ICSE Solutions for Class 9 Chemistry – Study of Gas Laws

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Frank ICSE Solutions for Class 9 Chemistry – Study of Gas Laws

PAGE NO :21
Solution 1:
An ideal gas can be characterized by three state variables:

  1. Absolute pressure (P),
  2. Volume (V), and
  3. Absolute temperature (T).

Solution 2:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 1

Solution 3:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 2

Solution 4:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 19

Solution 5:
Kelvin zero is – 273.15 oC.

Solution 6:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 3

Solution 7:
The standard temperature and pressure (STP) by general convention are 0 oC(273 K) and 1 atm(760 mm Hg).

Solution 8:

  1. The value of standard temperature is (i) 0 oC and (ii) 273 K
  2. The value of standard pressure is (i) 1 atm, (ii) 760 mm of Hg, (iii)76 cm of Hg, (iv)760 torr

Solution 9:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 4

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 5
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 6

Solution 11:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 7

Solution 12:
There is simultaneous effect of temperature and pressure changes on the volume of a given mass of a gas. So, when stating the volume of a gas, the pressure and temperature should also be given.

PAGE NO :22
Solution 13:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 8

Solution 14:

  1. Volume of a gas would be reduced to zero at 0 K (-273 0C).All temperatures on the Kelvin scale are positive, so Kelvin scale has been adopted for chemical calculation.
  2. At absolute zero temperature, volume of a gas would be reduced to zero. Theoretically,this is the lowest temperature that can be reached. At this temperature all molecular motions cease. Thus, practically this temperature is impossible to attain because on cooling gases liquefy and Charles’ law is no more applicable.
  3. According to combined gas law equation, there is simultaneous effect of temperature and pressure changes on the volume of a given mass of a gas. So, when stating the volume of a gas, the pressure and temperature should also be given.

Solution 15:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 9

Solution 16:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 10

Solution 17:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 11

Solution 18:

  1. True
  2. False
  3. False
  4. False
  5. False

Solution 19:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 12

Solution 20:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 13

Solution 21:
We trap a definite quantity of  air in the closed vessel. At any point, the pressure on the air is equal to the atmospheric pressure plus the pressure due to the excess mercury column in the open end tube. By pouring mercury in the tube, we increase the pressure on the air and measure its volume under that pressure. We thus obtain a set of data for the volume of a fixed mass of air under different pressures.
For a given mass of air at constant temperature, the following observations are made-

  1. The volume of air decreases with increasing pressure and vice versa.
  2. The proportion by which the volume decreases or increases is the same by which the pressure increases or decreases.

Solution 22:

  1. Pressure will also be doubled.
  2. Pressure will be double.

Solution 23:

  1. 273
  2. absolute zero
  3. absolute temperature
  4. the average kinetic energy

Solution 24:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 14

Solution 25:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 15

Solution 26:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 16

Solution 27:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 17

Solution 28:
As weather balloon go higher into the atmosphere, the air becomes less dense, so air pressure drops. Because of this, the air that is already inside the balloon  expands to cope with the difference in pressure. The end result is that the balloon expands making it larger.

Solution 29:
Frank ICSE Solutions for Class 9 Chemistry - Study of Gas Laws 18

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Frank ICSE Solutions for Class 9 Chemistry – Matter and its Composition: Law of Conservation of Mass

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Frank ICSE Solutions for Class 9 Chemistry – Matter and its Composition: Law of Conservation of Mass

PAGE NO :8
Solution 1:
Matter is anything around us which occupies space and has mass. Example- Coal, Copper, Water, Oxygen, Kerosene
Key: Matter occupies space and has mass.

Solution 2:
The conditions for something to be called matter are –

  1. It should occupy space.
  2. It should have mass.

Solution 3:
Light and sound are not considered to be matter because they neither have mass nor do they occupy space.

PAGE NO :9
Solution 4:
Solid:

  1. The particles are held together by strong intermolecular forces and have minimum intermolecular space.
  2. Solids have definite mass, shape and volume.

Liquid:

  1. The intermolecular forces of liquid molecules are intermediate of molecules  of solids and gases while intermolecular spaces are greater than in solids.
  2. Liquids have definite mass and volume but not definite shape. They take the shape of the container.

Gas:

  1. The particles are held together by very weak intermolecular forces while intermolecular spaces are much greater than in solids.
  2. Gases have definite mass but not definite shape and volume. They take the shape of the container.

Solution 5:
Two reasons for saying that wood is a solid are-

  1. Wood has definite mass and shape.
  2. Their intermolecular forces are very strong so they cannot flow.

Solution 6:
The particles of gases are separated from each other by large spaces and intermolecular forces of attraction are the weakest in gases. They have least density. So, they can flow easily. Hence, gases have no fixed shape and volume.
Key: Intermolecular forces of attraction are the weakest in gases.

Solution 7:

  1. Gases
  2. Solid
  3. Solid

Solution 8:

Properties Solids Liquids Gases
1.State of packing The particles are closely packed and their positions are also fixed. The particles are loosely packed and their positions are not fixed. The molecules are wide apart and their positions are also not fixed.
2.Energy associated Particles can vibrate only to and fro about their mean positions. Therefore, they have small kinetic energy due to their motion. The particles can move about more freely and have considerable kinetic energy due to their motion. The particles move about freely and have maximum kinetic energy due to their motion.
3.Intermolecular forces The particles are held together by strong intermolecular forces. The particles are held together by weak intermolecular forces. The particles are held together by very weak intermolecular forces.
4.Physical features Solid has a crystalline structure with both definite size and definite shape. Liquid does not have a definite shape but has definite volume and can flow from higher to lower level. Gas has neither definite shape nor a definite volume but can flow and is easily compressible.

Solution 9:

  1. gases
  2. strong
  3. sublimation
  4. Condensation
  5. gaseous

Solution 10:
The postulates of the kinetic theory of matter-

  1. Composition of Matter: Matter, whether in the solid, liquid or gaseous state, is composed of very small particles which may be molecules, atoms or ions.
  2. Arrangement of Particles: These particles have spaces lying between them and these spaces are referred to as intermolecular spaces or interparticle spaces.
  3. Forces of Attraction: The forces of attraction between the molecules of a given substance are called intermolecular forces. The magnitude of this force depends upon the state of the substance and on the magnitude of the intermolecular spaces. As the intermolecular space increases, the intermolecular force decreases.
  4. Motion of the constituent particles: The particles are always in a state of motion. In solids, they vibrate about their mean positions and in liquids and gases, they move randomly.
  5. The kinetic energy of the particles increases with rise in temperature. As the temperature is increased, the particles undergo motion more vigorously and more  randomly.

Solution 11:

  1. Solid to liquid state: The conversion of a substance from the solid state to liquid state at a particular temperature is called melting or fusion. The heat energy supplied to the solid is absorbed by its molecules to gain kinetic energy. The kinetic energy increases the rate of vibration of the molecules. The force of attraction thus no longer holds the molecules close together and the solid gets change into liquid.
  2. Gas to liquid: The conversion of a substance from the gaseous state to its liquid state at a particular temperature is called condensation or liquefaction. On cooling, the gas molecules loose their kinetic energy in the form of lost heat and so molecular motion slows down. Decreased molecular motion causes a decrease in intermolecular space. The molecules come very close and the force of attraction between them correspondingly increases and the gas gets change into liquid.
  3. Liquid to gaseous state: The heat energy supplied to the liquid is absorbed by its molecules to gain kinetic energy and therefore the molecules move faster. This increases the intermolecular space. The intermolecular force of attraction decreases and liquid changes into gaseous state.
  4. Liquid to solid state: The conversion of a substance from the liquid state to solid state by cooling is called freezing. On cooling a liquid, the kinetic energy of the molecule is decreased. Due to decreased kinetic energy, the molecules cool down and come closer, thus reducing the intermolecular spaces. The force of attraction between the molecules thus increases. Now, the molecules are no longer in a position to be free or to migrate and liquid changes into a solid.

Solution 12:

  1. Freezing: The process of changing a liquid into a solid by cooling is called        freezing. Freezing means solidification. It occurs at a definite temperature called freezing point.
  2. Evaporation: The phenomenon involving the change of a substance from the liquid state to the gaseous state at room temperature or at any other temperature below its boiling point is called vaporization or evaporation.
  3. Boiling point: The temperature at which a liquid boils and changes rapidly into a gas at atmospheric pressure is called boiling point of the liquid.
  4. Melting point: The temperature at which a solid substance changes into its liquid state at 1 atmospheric pressure is called the melting point of that substance.

Solution 13:

  1. Size of naphthalene balls decreases – Sublimation
  2. Drying of wet clothes – Evaporation
  3. Wax melts in the sun – Melting
  4. Formation of clouds – Evaporation and Condensation

Solution 14:
Three compounds which are sublimate are-

  1. Camphor
  2. Naphthalene
  3. Iodine

Solution 15:
‘States of matter triangle’ shows inter-conversion of states of matters.
Frank ICSE Solutions for Class 9 Chemistry - Matter and its Composition Law of Conservation of Mass 1

Solution 16:
The process by which a liquid slowly converts into vapour state at a temperature below its boiling point is called evaporation.
The heat energy is absorbed by the liquid to convert it into vapour state. So, loss of heat causes cooling.

Solution 17:

  1. True
  2. True
  3. False
  4. True

Solution 18:
“In all physical and chemical changes, the total mass of the reactants is equal to that of the products”. So, in other words matter can neither be created nor destroyed.

Solution 19:
Frank ICSE Solutions for Class 9 Chemistry - Matter and its Composition Law of Conservation of Mass 2

Solution 20:

  1. Matter
  2. Solid
  3. Gas
  4. Solid
  5. Gas
  6. Gas
  7. Liquid
  8. Fluid
  9. Melting
  10. Vaporization

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Selina ICSE Solutions for Class 10 Maths – Compound Interest (Using Formula)

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Selina ICSE Solutions for Class 10 Maths – Compound Interest (Using Formula)

Selina ICSE Solutions for Class 10 Maths Chapter 2 Compound Interest (Using Formula)

Exercise 2(A)

Solution 1:
Given : P = ₹ 12,000; n = 3 years and r = 5%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q1
= ₹ 13,891.50
C.I. = ₹ 13,891.50 – ₹ 12,000
= ₹ 1,891.50

Solution 2:
Given : P = ₹ 15,000; n = 2 years ; r1 = 8%; r2 = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q2
= ₹ 17,820

Solution 3:
Given : P = ₹ 6,000; n = 3 years ; r1 = 5%; r2 = 8% and r3 = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q3
= ₹ 7,484.40
C.I. = ₹ 7,484.40 – ₹ 6,000 = ₹ 1,484.40

Solution 4:
Given : Amount = ₹ 5,445; n = 2 years and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q4

Solution 5:
Given : C.I = ₹ 768.75; n = 2 years and r = 5%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q5

Solution 6:
Given : C.I = ₹ 1,655; n = 3 years and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q6

Solution 7:
Given : Amount = ₹ 9,856; n = 2 years ; r1 = 10%; r2 = 12%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q7

Solution 8:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q8
The sum is ₹ 16,000

Solution 9:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q9
At 5% per annum the sum of ₹ 6,000 amounts to ₹ 6,615 in 2 years when the interest is compounded annually.

Solution 10:
Let Principal = ₹ y
Then Amount = ₹ 1.44y
n = 2 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q10

Solution 11:
Given : P = ₹ 18,000; C.I. = ₹ 5,958 and n = 3 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q11

Solution 12:
Given: P = ₹ 5,000; A = ₹ 6,272 and n = 2 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q12

Solution 13:
Given : P = ₹ 7,000; A = ₹ 9,317 and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q13

Solution 14:
Given : P = ₹ 4,000; C.I. = ₹ 630.50 and r = 5%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q14

Solution 15:
Let share of A = ₹ y
share of B = ₹ (28,730 – y)
rate of interest = 10%
According to question
Amount of A in 3 years = Amount of B in 5 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q15
Therefore share of A = ₹ 15,730
Share of B = ₹ 28,730 – ₹ 15,730 = ₹ 13,000

Solution 16:
Let share of Rohit = ₹ y
share of Rajesh = ₹ (34,522 – y)
rate of interest = 5%
According to question
Amount of Rohit in 12 years = Amount of Rajesh in 9 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q16
Therefore share of Rohit = ₹ 16,000
Share of Rajesh = ₹ 34,522 – ₹ 16,000 = ₹ 18,522

Solution 17:
(i) Let share of John = ₹ y
share of Smith = ₹ (44,200 – y)
rate of interest = 10%
According to question
Amount of John in 4years = Amount of Smith in 2years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q17-i
Therefore share of John = ₹ 20,000
Share of Smith = ₹ 44,200- ₹ 20,000 = ₹ 24,200
(ii) Amount that each will receive
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q17-ii

Solution 18:
The amount of money in the account = ₹ 22,000
Compound interest for the first year = Simple interest for the first year
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q18

Solution 19:
Let’s ₹ x be the sum of the money.
Let A1 be the amount obtained at the end of the 1st year.
Let A2 be the amount obtained at the end of the 2nd year.
Let R be the rate of interest.
The amounts of are in the ratio 20:21.
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q19
But rate on interest cannot be negative hence R = 5.
Therefore the rate of interest is 5%.

Solution 20:
Let’s ₹ x be the sum of the money.
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2a-q20
The sum of the money is ₹ 30,000.

Exercise 2(B)

Solution 1:
Given: P = ₹ 7,400; r = 5% p.a. and n = 1 year
Since the interest is compounded half-yearly,
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q1

Solution 2:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q2

Solution 3:
For the first 2 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q3-i
Amount in the account at the end of the two years is ₹ 22,400.
For the remaining one year
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q3-ii
The total amount to be paid at the end of the three years is ₹ 27,104.

Solution 4:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q4
The sum of ₹ 24,000 amount ₹ 27,783 in one and a half years at 10% per annum compounded half yearly.

Solution 5:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q5

Solution 6:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q6
The rate of interest is 8%.

Solution 7:
Given: P = ₹ 1,500; C.I.= ₹ 496.50 and r = 20%
Since interest is compounded semi-annually
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q7

Solution 8:
Given: P = ₹ 3,500; r = 6% and n = 3 years
Since interest is being compounded half-yearly
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q8

Solution 9:
Given: P = ₹ 12,000; n = 1 ½ years and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q9-i
To calculate C.I.
For 1 year
P = ₹ 12,000; n = 1 year and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q9-ii
For next ½ year
P = ₹ 13,200; n = ½ year and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q9-iii
∴ C.I. = ₹ 13,860 – ₹ 12,000 = ₹ 1,860
∴ Difference between C.I. and S.I = ₹ 1,860 – ₹ 1,800 = ₹ 60

Solution 10:
Given: P = ₹ 12,000; n = 1 ½ years and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q10-i
To calculate C.I.(compounded half-yearly)
P = ₹ 12,000; n = 1 ½ years and r = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q10-ii
∴ C.I. = ₹ 13,891.50 – ₹ 12,000 = ₹ 1,891.50
∴ Difference between C.I. and S.I
= ₹ 1,891.50 – ₹ 1,800 = ₹ 91.50

Solution 11:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q11

Solution 12:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2b-q12

Exercise 2(C)

Solution 1:
Initial height(P)= 80 cm
Growth rate = 20%
∴ Growth after 3 months
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q1

Solution 2:
Cost of machine in 2008 = ₹ 44,000
Depreciation rate = 12%
(i) Cost of machine at the end of 2009
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q2-i
(ii) Cost of machine at the beginning of 2007(P)
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q2-ii

Solution 3:
Value of a machine at the end of 2004(P)= ₹ 27,000
Value of a machine at the beginning of 2007(A)= ₹ 21,870
Time(n)= 2 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q3-i
(ii) The value of machine at the beginning of 2004(P)
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q3-ii

Solution 4:
Let x be the value of the article.
The value of an article decreases for two years at the rate of 10% per year.
The value of the article at the end of the 1st year is
x – 10% of x = 0.90x
The value of the article at the end of the 2nd year is
0.90x – 10% of (0.90x) = 0.81x
The value of the article increases in the 3rd year by 10%.
The value of the article at the end of 3rd  year is
0.81x + 10% of (0.81x) = 0.891x
0.891x = 40,095
⇒ x = 45,000
The value of the article at the end of 3 years is ₹ 40,095.
The original value of the article is ₹ 45,000.

Solution 5:
Population in 2005(P) = 64,000
Let after n years its population be 74,088(A)
Growth rate = 5% per annum
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q5

Solution 6:
Let the population in the beginning of 1998 = P
The population at the end of 1999 = 2,85,120(A)
r1 = – 12% and r2 = +8%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q6

Solution 7:
Let sum of money be Rs P and rate of interest= r%
Money after 1 year = ₹ 16,500
Money after 3 years = ₹ 19,965
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q7

Solution 8:
Given: P = ₹ 7,500 and Time(n) = 2 years
Let rate of interest = y%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q8

Solution 9:
Let Principal be Rs y and rate= r%
According to 1st condition
Amount in 10 years = Rs 3y
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q9

Solution 10:
At the end of the two years the amount is
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q10-i
Mr. Sharma paid ₹ 19,360 at the end of the second year.
So for the third year the principal is A1 – 19,360.
Also he cleared the debt by paying ₹ 31,944 at the end of the third year.
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q10-ii
Mr. Sharma borrowed ₹ 40,000.

Solution 11:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q11

Solution 12:
Let ₹ x and ₹ y be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of 25 years.
Pramod will attain the age of 25 years after 25 – 16 = 9 years
Rohit will attain the age of 25 years after 25 -18 = 7 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2c-q12
Pramod and Rohit should invest in 400:441 ratio respectively such that they will get the same sum on attaining the age of 25 years.

Exercise 2(D)

Solution 1:
Let ₹ 100 be the principal amount.
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q1
When the principal is ₹ 100 the compound interest is ₹ 10.25.
Therefore the effective rate per annum is 10.25%.

Solution 2:
Let ₹ x be the value of the property.
Value of the property decreases at the rate of 6 1/4 percent at the beginning of that year.
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q2-i
Value of the property at the end of the 2nd year = Value of the property at the beginning of the 3rd year
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q2-ii
Value of the property at the beginning of 2nd year = value of property at the beginning of 1st year
Value of property
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q2-iii
Value of the property at the beginning of 2 years is ₹ 2,56,000.

Solution 3:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q3

Solution 4:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q4

Solution 5:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q5

Solution 6:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q6

Solution 7:
(i) Present value of machine(P) = ₹ 97,200
Depreciation rate = 10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q7-i
= ₹ 78732
(ii) Present value of machine(A) = ₹ 97,200
Depreciation rate = 10% and time = 2 years
To calculate the cost 2 years ago
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q7-ii

Solution 8:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q8

Solution 9:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q9

Solution 10:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q10

Solution 11:
Given: C.I. for the 2nd year = ₹ 4,950 and rate = 15%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q11-i
Then amount at the end of 2nd year= ₹ 33,000
For first 2 years
A = ₹ 33,000; r1 =10%
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q11-ii

Solution 12:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q12

Solution 13:
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q13

Solution 14:
P = ₹ 60,000, R = 5%, n = 2 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q14-i
The money lender deducts the interest that would be due at the end of the period and handed over the balance to Mrs. Shukla.
Amount received by Mrs. Shukla = ₹ 60,000 – ₹ 6,000 = ₹ 54,000
Now sum deposited in the bank (P) = ₹ 54,000
R = 5% compounded annually, n = 2 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q14-ii
The amount of money that Mrs. Shukla will have to add to pay the money back to the money lender
= ₹ 60,000 – ₹ 59,535 = ₹ 465

Solution 15:
Let ₹ x be the sum of money.
Rate = 5 % p.a. Simple interest = ₹ 1,200, n = 3years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q15-i
The amount due and the compound interest on this sum of money at the same rate and after 2 yers
P = ₹ 8,000; rate = 5% p.a., n = 3 years
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q15-ii
The amount due after 2 years is ₹ 8,820 and the compound interest is ₹ 820.

Solution 16:
Let x% be the rate of interest.
P = ₹ 6,000, n = 2 years, A = ₹ 6,720
(i) For the first year
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q16-i
The rate of interest is x% = 12%.
(ii) The amount at the end of the second year.
selina-icse-solutions-class-10-maths-compound-interest-using-formula-ex-2d-q16-ii
The amount at the end of the second year = ₹ 7,526.40

Selina ICSE Solutions for Class 10 Maths – Compound Interest (Without using formula)

by

Selina ICSE Solutions for Class 10 Maths – Compound Interest (Without using formula)

Selina ICSE Solutions for Class 10 Maths Chapter 1 Compound Interest (Without using formula)

Exercise 1(A)

Solution 1:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q1

Solution 2:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q2

Solution 3:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q3

Solution 4:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q4

Solution 5:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q5

Solution 6:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q6

Solution 7:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q7

Solution 8:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q8

Solution 9:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q9

Solution 10:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q10

Solution 11:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q11

Solution 12:
Let Rs.x be the sum.
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q12-i
Compound interest
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q12-ii
The difference between the simple interest and compound interest at the rate of 8% per annum compounded annually should be ₹ 64 in 2  years.
⇒ ₹ 0.08x – ₹ 0.0864x = ₹ 64
⇒ ₹ 0.0064x = ₹ 64
⇒ x = ₹ 10000
Hence the sum is ₹ 10000.

Solution 13:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q13

Solution 14:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q14

Solution 15:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q15

Solution 16:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q16

Solution 17:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q17

Solution 18:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q18

Solution 19:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q19
Amount = ₹ 30,400 + ₹ 3,040 + ₹ 4,000 = ₹ 37,440
The amount in Mrs. Kapoor’s account on 01/01/2012 is ₹ 37,440.

Solution 20:
(i) Let x% be the rate of interest charged.
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1a-q20
The compound interest for the second year is ₹ 920
Rs. (80x + 1.20x2) = ₹ 920
⇒ 1.20x2 + 80x – 920 = 0
⇒ 3x2 + 200x – 2300 = 0
⇒ 3x2 + 230x – 30x – 2300 = 0
⇒ x(3x + 230) -10(3x + 230) = 0
⇒ (3x + 230)(x – 10) = 0
⇒ x = -230/3 or x = 10
As rate of interest cannot be negative so x = 10.
Therefore the rate of interest charged is 10%.
(ii) For 1st year:
Interest = ₹ 120x = ₹ 1200
For 2nd year:
Interest = ₹ (80x + 1.20x2) = ₹ 920
The amount of debt at the end of the second year is equal to the addition of principal of the second year and interest for the two years.
Debt = ₹ 8,000 + ₹ 1200 + Rs.920 = ₹ 10,120

Exercise 1(B)

Solution 1:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q1

Solution 2:
Difference between the C.I. of two successive half-years
= ₹ 760.50 – ₹ 650 = ₹ 110.50
₹ 110.50 is the interest of one half-year on ₹ 650
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q2

Solution 3:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q3

Solution 4:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q4

Solution 5:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q5

Solution 6:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q6

Solution 7:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q7

Solution 8:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q8

Solution 9:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q9

Solution 10:
(i) The interest charged is compounded because if the interest charged is simple, then the interest for two years will be double of interest for one year which is not given.
(ii) C.I. for 1st year= ₹ 720
C.I. for two years= ₹ 1,497.60
C.I. for 2nd year = ₹ 1,497.60 – ₹ 720 = ₹ 777.60
Difference between the C.I. of two successive years
= ₹ 777.60 – ₹ 720
= ₹ 57.60
₹ 57.60 is the interest for one year on ₹ 720.
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q10

Solution 11:
(i) C.I. for second year = ₹ 864
C.I. for third year = ₹ 933.12
Difference between the C.I. of two successive years= ₹ 933.12 – ₹ 864= ₹ 69.12
₹ 69.12 is the interest of one year on ₹ 864
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q11-i
(ii) Let the sum of money= ₹ 100
Interest on it for 1st year= 8% of ₹ 100= ₹ 8
Amount in one year= ₹ 100+ ₹ 8= ₹ 108
Similarly, C.I. for 2nd year= 8% of ₹ 108 = ₹ 8.64
When C.I. for 2nd year is ₹ 8.64, sum = ₹ 100
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q11-ii
Principal for 4th year= ₹ 10,000+₹ 800+₹ 864+₹ 933.12 = ₹ 12,597.12
Interest for 4th year= 8% of ₹ 12,597.12 = ₹ 1,007.77

Solution 12:
(i) Amount in three years = ₹ 20,160
Amount in four years = ₹ 24,192
Difference between the amounts of two successive years
= ₹ 24,192 – ₹ 20,160= ₹ 4,032
⇒ ₹ 4,032 is the interest of one year on ₹ 20,160
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q12-i
(ii) Let amount in two years = ₹ 100
And amount in three years = ₹ 100+ 20% of ₹ 100
= ₹ 100+ ₹ 20
= ₹ 120
When amount in 3 years is ₹ 120, amount in two years = ₹ 100
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q12-ii
(iii) Amount in 5 years = ₹ 24,192+ 20% of ₹ 24,192
= ₹ 24,192 +₹ 4,838.40
= ₹ 29,030.40
Solution 13:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q13
(ii)The total interest paid in two years = ₹ 350 + ₹ 560 = ₹ 910
(iii) The total amount of money paid in two years to clear the debt
= ₹ 8,000+ ₹ 910
= ₹ 8,910

Solution 14:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q14

Solution 15:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q15

Solution 16:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q16

Solution 17:
(i) Difference between depreciation in value between the first and second years = ₹ 4,000 – ₹ 3,600 = ₹ 400
⇒ Depreciation of one year on ₹ 4,000 = ₹ 400
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q17-i
(ii) Let ₹ 100 be the original cost of the machine.
Depreciation during the 1st year = 10% of ₹ 100 = ₹ 10
When the values depreciates by ₹ 10 during the 1st year, Original cost = ₹ 100
⇒ When the depreciation during 1st year = ₹ 4,000
The original cost of the machine is ₹ 40,000.
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1b-q17-ii
(iii) Total depreciation during all the three years
= Depreciation in value during(1st year + 2nd year + 3rd year)
= ₹ 4,000 + ₹ 3,600 + 10% of (₹ 40,000 – ₹ 7,600)
= ₹ 4,000 + ₹ 3,600 + ₹ 3,240
= ₹ 10,840
The cost of the machine at the end of the third year
= ₹ 40,000 – ₹ 10,840 = ₹ 29,160.

Exercise 1(C)

Solution 1:
Let the sum of money be ₹ 100
Rate of interest = 10% p.a.
Interest at the end of 1st year = 10% of ₹ 100 = ₹ 10
Amount at the end of 1st year = ₹ 100 + ₹ 10 = ₹ 110
Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11
Amount at the end of 2nd year = ₹ 110 + ₹ 11 = ₹ 121
Interest at the end of 3rd year = 10% of ₹ 121 = ₹ 12.10
Sum of interest of 1st year and 3rd year = ₹ 10 + ₹ 12.10 = ₹ 22.10
When sum of both interest is ₹ 22.10, principal is ₹ 100
When sum of both interest is ₹ 1,768, principal = \(\frac { 100\times 1768 }{ 22.10 }\) ₹ = ₹ 8,000

Solution 2:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1c-q2

Solution 3:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1c-q3

Solution 4:
Cost of machine= ₹ 32,000
Depreciation rate every year = 5%
∴ Cost of machine after one year = ₹ 32,000- 5% of ₹ 32,000
= ₹ 32,000- ₹ 1,600
= ₹ 30,400
Cost of machine after two year = ₹ 30,400- 5% of ₹ 30,400
= ₹ 30,400- ₹ 1,520
= ₹ 28,880
∴ Total depreciation in two years = ₹ 32,000 – ₹ 28,880 = ₹ 3,120.

Solution 5:
Let the sum of money be ₹ 100.
Rate of interest= 10%p.a.
Interest at the end of 1st year = 10% of ₹ 100= ₹ 10
Amount at the end of 1st year = ₹ 100 + ₹ 10= ₹ 110
Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11
Amount at the end of 2nd year = ₹ 110 + ₹ 11= ₹ 121
Interest at the end of 3rd year = 10% of ₹ 121= ₹ 12.10
∴ Difference between interest of 3rd year and 1st year = ₹ 12.10 – ₹ 10 = ₹ 2.10
When difference is ₹ 2.10, principal is ₹ 100.
When difference is ₹ 252, principal = \(\frac { 100\times 252 }{ 2.10 }\) = ₹ 12,000.

Solution 6:
(i) C.I. for 2nd year = ₹ 9,680
C.I. for 3rd year = ₹ 10,648
Difference in both interests = ₹ 10,648 – ₹ 9,680 = ₹ 968
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1c-q6
(ii) Interest for 4th year = ₹ 10,648+ 10% of ₹ 10,648
= ₹ 10,648 + ₹ 1,064.80
= ₹ 11,712.80
(iii) Let principal be ₹ 100
Rate of interest= 10% p.a.
Interest at the end of 1st year = 10% of ₹ 100= ₹ 10
Amount at the end of 1st year = ₹ 100 + ₹ 10= ₹ 110
Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11
When C.I. for 2nd year is ₹ 11, principal is ₹ 100
When C.I. for 2nd year is ₹ 9,680, principal = ₹ \(\frac { 100\times 9680 }{ 11 }\) = ₹ 88,000
Interest for 1st year = 10% of ₹ 88,000 = ₹ 8,800.

Solution 7:
(i) Amount in two years = ₹ 9,680
Amount in three years = ₹ 10,648
∴ Difference in both amounts = ₹ 10,648 – ₹ 9,680 = ₹ 968
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1c-q7
(ii) Amount in 4 years = ₹ 10,648+ 10% of ₹ 10,648
= ₹ 10,648 + ₹ 1,064.80
= ₹ 11,712.80
(iii) Let principal be ₹ 100
Rate of interest= 10%p.a.
Interest at the end of 1st year = 10% of ₹ 100= ₹ 10
Amount at the end of 1st year = ₹ 100 + ₹ 10= ₹ 110
Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11
Amount at the end of 2nd year = ₹ 110 +₹ 11= ₹ 121
When amount at the end of 2nd year is ₹ 121, principal is ₹ 100
When amount at the end of 2nd year is ₹ 9,680, principal
= ₹ \(\frac { 100\times 9680 }{ 121 }\)
= ₹ 8,000
∴ Amount in one year = ₹ 8,000+10% of ₹ 8,000
= ₹ 8,000 + ₹ 800
= ₹ 8,800

Solution 8:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1c-q8

Solution 9:
selina-icse-solutions-class-10-maths-compound-interest-without-using-formula-ex-1c-q9

Solution 10:
(i) The population of a town increases by 10% every 3 years.
The population of the town after 3 years
= 72,600 + 10% of 72,600
= 72,600 + 7,260
= 79,860
The population of the tower after 6 years
= 79,860 + 10% of 79,860
= 79,860 + 7,986
= 87,846
The population of the town after 6 years is 87,846.
(ii) Let x be the population of the town 6 years ago.
The present population of the town is 72,600.
The population of the town 3 years ago
= x + 10% of x
= x + 0.10x
= 1.10x
The present population of the town
= 1.10x + 10% of 1.10x
⇒ 72,600 = 1.10x + 0.110x
⇒ 72,600 = 1.210x
⇒ x = 60,000
The population of the town before 6 years ago was 60,000.

ICSE Solutions for Class 10 Chemistry – Practical Chemistry

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ICSE Solutions for Class 10 Chemistry – Practical Chemistry

ICSE SolutionsSelina ICSE Solutions

APlusTopper.com provides ICSE Solutions for Class 10 Chemistry Chapter 12 Practical Chemistry for ICSE Board Examinations. We provide step by step Solutions for ICSE Chemistry Class 10 Solutions Pdf. You can download the Class 10 Chemistry ICSE Textbook Solutions with Free PDF download option.

Download Formulae Handbook For ICSE Class 9 and 10

Short Questions

Question 1: Name the three ions which can be identified by K4[Fe(CN)6] solution.
Answer: Fe3+, Z2+, Cu2+

Question 2: State the colour of the following :
(i) Lead monoxide (h) Hat zinc oxide
(iii) Copper carbonate (iv) Copper nitrate crystals
Answer: (i) Lime yellow (ii) Yellow
(iii) Emerald green (iv) Deep blue

Question 3: Mention the colour changes observed when the following indicators are added to acids :
(i) Alkaline phenolphthalein solution (ii) Methyl orange solution
(iii) Neutral litmus solution
Answer: (i) From pink to colourless (ii) From orange to pink (red)
(iii) From colourless to red

Question 4: Identify the anion present in the following compounds :
(i) Compound X on heating, with copper turnings and concents sulphuric acid liberates a reddish brown gas.
(ii) When a solution of compound Y is treated with silver nitrate solution a white precipitate is obtained which is soluble in excess of ammonium hydroxide solution.
(iii) Compound Z which on reacting with dilute sulphuric acid liberates a gas which lime lune water milky, but the gas has no effect on acidified potassium dichromate solution.
(iv) Compound L on reacting with barium chloride solution gives a white precipitate insoluble in dilute hydrochloric acid or dilute nitric acid.
Answer: (i) Nitrate ion, NO3– (ii) Chloride ion, Cl–
(iii) Carbonate ion, CO3– (iv) Sulphate ion, SO42-

Question 5: Identify the substances P, Q, R, S and T in each case based on the information given below:
(i) The deliquescent salt P, turns yellow on dissolving in water, and gives a reddish brown precipitate with sodium hydroxide solution.
(ii) The white crystalline solid Q is soluble in water. It liberates a pungent smelling gas when heated with sodium hydroxide solution.
(iii) The pale green solid R turns reddish brown on heating. Its aqueous solution gives a white precipitate with barium chloride solution. The precipitate is insoluble in mineral acids.
(iv) The reddish brown liquid S is dissolved in water. When Ethyne gas is passed through it, turns colourless.
Answer: (i) Hydrogen chloride gas (ii) Nitric oxide (iii) Oxygen (iv) Ammonia

Question 6: Select from the list given (A to E) one substance in each case which matches the description given in parts (i) to (v). (Note: Each substance is used only once in the answer.)
(A) Nitroso Iron(II) sulphate (B) Iron(III) Chloride (C) Chromium sulphate (D) Lead(II) Chloride (E) Sodium chloride.
(i) A compound which is deliquescent.
(ii) A compound which is insoluble in cold water, but soluble in hot water.
(iii) The compound responsible for the brown ring during the brown ring test of nitrate ion.
(iv) A compound whose aqueous solution is neutral in nature.
(v) The compound which is responsible for the green colouration when sulphur dioxide is passed through acidified potassium dichromate solution.
Answer: (i) B, Iron (III) chloride
(ii) D, Lead (II) chloride
(iii) A, Nitroso Iron (II) sulphate
(iv) E, Sodium chloride
(v) C, Chromium sulphate

Question 7: The questions (i) to (v) refer to the following salt solutions listed A to F:
A. Copper nitrate B. Iron (II) sulphate.
C. Iron (III) chloride D. Lead nitrate
E. Magnesium sulphate F. Zinc chloride.
(i) Which two solutions will give a white precipitate when treated with dilute Hydrochloric acid followed by Barium chloride solution ?
(ii) Which two solutions will give a white precipitate when treated with dilute Nitric acid followed by Silver nitrate solution ?
(iii) Which solution will give a white precipitate when either dilute Hydrochloric acid or dilute Sulphuric acid is added to it ?
(iv) Which solution becomes a deep /inky blue colour when excess of Ammonium hydroxide is added to it ?
(v) Which solution gives a white precipitate with excess Ammonium hydroxide solution ?
Answer: (i) B and E (Iron II sulphate and magnesium sulphate).
(ii) C and F (Iron III chloride and zinc chloride)
(iii) D (lead nitrate)
(iv) A (copper nitrate)
(v) F (zinc chloride)

Question 8: A white crystalline solid A on dissolving in water forms a neutral solution. When solid A is heated with calcium hydroxide, it gives off colourless gas B having a sharp biting smell. The gas fumes very strongly in air and turns red litmus blue. The aqueous solution of A on treating with lead acetate solution forms a white precipitate C. The precipitate is insoluble in all acids.
(i) Name the cation present in solid A.
(ii) Name the anion present in solid A.
(iii) Name the solid A and write its chemical formula.
(iv) Name the colourless gas B.
(v) Write one more chemical test for the identification of gas B.
(vi) Name the white precipitate C.
(vii) Write fully balanced chemical equations for:
(a) Solid A and calcium hydroxide.
(b) Solution of solid A and lead acetate solution.
Answer: (i) Cation in A is ammonium (NH4+)
(ii) Anion in A is sulphate (SO42- )
(iii) Solid A is ammonium sulphate [(NH4)2SO4]2
(iv) The colourless gas B is ammonia gas.
(v) Bring a rod dipped in HCl sol. near the gas. The dense white fumes of ammonium chloride are formed.
(vi) White precipitate C is lead sulphate.
(vii)
(a) (NH4)2SO4 + Ca(OH)2   ⟶  CaSO4 + 2NH3 + 2H2O
(b) (NH4)2SO4 + (CH3COO)2Pb  ⟶  PbSO4 + 2CH3COONH4

Question 9: Identify the following solids.
(i) A silvery white solid which floats on the surface of water and reacts violently to give tiny bubbles of a colourless gas. The gas burns in air with a pop sound. The chloride of solid gives a non-persistent lilac colour to non-luminous bunsen flame.
(ii) A pale yellow compound gives a persistent golden yellow colour in non-luminous bunsen burner flame. When treated with water, it liberates a colourless gas, which burns with a pop sound.
(iii) A white solid on treating with water gives off a colourless gas which has a sharp biting smell and turns red- litmus blue. The white solid formed after the reaction is soluble in cone, sodium hydroxide solution.
Answer: (i) The solid is potassium metal (ii) The solid is sodium hydride
(iii) The solid is aluminium nitride

Question 10: Identify the cations and anions in each case and write relevant equations wherever necessary.
(i) A white crystalline solid on heating swells and gives off colourless vapours, which condense to form a colourless liquid X. The X turns anhydrous copper sulphate solution blue. The flame test shows the flame as persistent golden yellow. When the solution of solid is treated with HCl, it gives off a colourless gas Y which turns lime water milky.
(ii) A blue solid gives a non-persistent green flame during the flame test. When the solid is heated strongly, it leaves behind a white residue. The residue on treating with few drops of water changes to blue colour.
(iii) A white solid gives a non-persistent brick red flame. On treating with dilute sulphuric acid, it gives off a colourless gas which turns acidified potassium dichromate solution green.
(iv) A white solid gives a non-persistent black colour. On treating with dilute sulphuric acid, it gives off a highly offensive smell. The gas turns lead acetate paper black.
Answer:
ICSE Solutions for Class 10 Chemistry - Practical Chemistry 1

Figure/Table Based Questions

Question 1: Salts A, B, C, D and E undergo reactions (i) to (v) respectively. Identify the anion present in these salts on the basis of these reactions. Tabulate your answers in the format given below :
(i) When silver nitrate solution is added to a solution of A, a white precipitate, insoluble in dilute nitric acid, is formed.
(ii) Addition of dilute hydrochloric acid to B produces a gas which turns lead acetate paper black.
(iii) When a freshly prepared solution of ferrous sulphate is added to a solution of C and concentrated sulphuric acid is gently poured from the side of the test-tube, a brown ring is formed.
(iv) When dilute sulphuric acid is added to D, a gas is produced which turns acidified potassium dichromate solution from orange to green.
(v) Addition of dilute hydrochloric acid to E produces an effervescence. The gas produced turns lime water milky but does not affect acidified potassium dichromate solution.

Salt Anion
A
B
C
D
E

Answer:
(i) A—Chloride,
(ii) B—Sulphide,
(iii) C—Nitrate,
(iv) D—Sulphite,
(v) E—Carbonate.

Question 2: Sodium hydroxide solution is added first in a small quantity, then in excess to the aqueous salt solutions of copper (II) sulphate, zinc nitrate, lead nitrate, calcium chloride and iron (III) sulphate. Copy the following table and write the colour of the precipitate in (i) to (v) and the nature of the precipitate (soluble or insoluble) in (vi) to (x).

Aqueous salt solution Colour of precipitate when NaOH is added in a small quantity Nature of precipitate (soluble or insoluble) when NaOH is added in excess
Copper (II) sulphate (i) (vi)
Zinc nitrate (ii) (vii)
Lead nitrate (iii) (viii)
Calcium chloride (iv) (ix)
Iron (III) sulphate (v) (x)

Answer:

Aqueous salt solution Colour of precipitate when NaOH is added in a small quantity. Nature of precipitate (soluble or insoluble) when NaOH is added in excess.
Copper (II) sulphate Blue Insoluble
Zinc nitrate White Soluble
Lead nitrate White Soluble
Calcium chloride White Insoluble
Iron (El) sulphate Reddish Brown Insoluble

Chemical Tests

Question 1:

  1. Zinc sulphate solution and Zinc chloride solution.
  2. Iron (II) chloride solution and Iron (III) chloride solution.
  3. Calcium nitrate solution and Calcium chloride solution.
  4. Sodium carbonate and sodium sulphite.
  5. Ferrous nitrate and Lead nitrate
  6. Manganese dioxide and Copper (II) oxide.
  7. Sodium chloride and sodium nitrate.
  8. Calcium nitrate and lead nitrate.
  9. Lead nitrate and zinc nitrate.
  10. Sodium sulphite anti sodium sulphate
  11. Ammonium chloride and ammonium nitrae.
  12. Zinc carbonate and lead carbQnate.
  13. Potassium chloride and sodium chloride.
  14. Powdered coke and manganese dioxide.
  15. Copper oxide and manganese dioxide.
  16. Washing soda and baking soda..
  17. Potassium nitrate and potassium nitrite.
  18. Mercuric oxide and red lead.
  19. Ferrous ions and ferric ions.

Answer:

  1. When BaCl2 solution is added to the given solution, ZnSO4 gives a white ppt while no ppt. is obtained with ZnCl2 solution.
  2. When NaOH solution is added to the given solution, Iron (II) chloride gives dirty green ppt. while reddish brown ppt. is obtained with Iron (III) chloride.
  3. When AgNO3 solution is added to the given solution, CaCl2 solution will give a white ppt. while no change is observed with calcium nitrate solution.
  4. Sodium carbonate when treated with dil H2SO4 will liberate colourless odourless gas with brisk effervescence. The gas will turn lime water milky and will have no effect on acidified potassium permanganate solution.
    Sodium sulphite when treated with dil H2SO4 will liberate a colourless gas having suffocating – smell of burning sulphur. The gas will turn acidified potassium permanganate solution from pink to colourless.
  5. Aqueous ferrous nitrate when treated with NaOH solution gives a dirty green ppt. Aqueous lead nitrate when treated with NaOH solution gives a white ppt. which dissolves in excess of NaOH.
  6. When manganese dioxide is heated with cone. HCl a pungent smelling greenish yellow chlorine gas is evolved. When copper oxide is heated with cone. HCl no such gas is evolved.
  7. On heating with cone. H2SO4 sodium chloride gives pungent colourless gas which turns silver nitrate solution curdy white, whereas sodium nitrate gives a brown gas which shows no effect on silver nitrate solution.
  8. Aqueous solution of calcium nitrate gives a white precipitate with sodium hydroxide which is insoluble in excess of sodium hydroxide solution. The aqueous solution of lead nitrate gives a white precipitate with caustic soda solution which dissolves in excess of canstic soda solution.
  9. Aqueous solution of lead nitrate gives white precipitate with ammounium hydroxide which is insoluble in excess ammonium hydroxide white the sodium of zinc nitrate gives a white precipitate which is soluble in excess ammonium hydroxide.
  10. Solution of both gives white precipitate with barium chloride solution. The precipitate of barium Sulphite is soluble in dil hydrochloric acid, white the precipitate of barium sulphite is insoluble in dil. hydrochloric acid.
  11. Heat each of salt strongly in a hard glass test tube. In case of ammonium chloride, sublimation takes place and white powdery deposits settles near the mouth of test tube. In case of ammonium nitrate, an explosion takes place and nitrous oxide and stream are given off. No residue is left in the test tube.
  12. Heat each of the carbonates strongly in a hard glass test tube. In case of zinc carbonate, residue is yellow when hot and white then cold. In case of lead carbonate, residue is reddish-brown when hot and yellow when cold.
  13. Make a thick paste of each salt in HCl. Take a small amount of the paste on clean platinum wire and introduce it in non-luminous bunsen flame. In case of potassium chloride, a non-persistent lilac flame is formed. In case of sodium chloride, a persistent golden yellow Marne is formed.
  14. Heat each of the susbtances with cone, hydrochloric add. In case of coke, no visible
    reaction takes place. In case of manganese dioxide, greenish-yellow gas chlorine is evolved.
  15. Warm each of the oxides with dilute hydrochloric acid. In case of copper oxide, a greenish-blue solution of copper chloride is formed. In case of manganese dioxide, no reaction takes place.
  16. Heat each of the salt strongly in a hard glass test tube and pass the gas evolved through lime-water. In case of washing soda, lime-water does not turn milky. In case of baking soda, lime-water turns milky.
  17. Treat each of the salts with dilute sulphuric acid and gently warm. In case of potassium nitrate, no visible reaction takes place. In case of potassium nitrite, reddish-brown gas, nitrogen dioxide is given off.
  18. Heat each of the oxides strongly in a hard glass test tube. In case of mercuric oxide, tiny droplets of mercury are seen near the mouth of the test tube and residue is black. In case of red lead, no tiny droplets are seen. The residue is reddish-brown when.hot and yellow when cold.
  19. Treat the solution of each ion with sodium hydroxdide solution. In case of ferrous ions, a dirty green ppt. of ferrous hydroxide is formed. In case of ferric ions, a reddish-brown ppt. of ferric hydroxide is formed.

For More Resources

ICSE Geography Question Paper 2015 Solved for Class 10

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ICSE Geography Previous Year Question Paper 2015 Solved for Class 10

ICSE Paper 2015
GEOGRAPHY

(Two hours)
Answers to this Paper must be written on the paper provided separately.
You will not be allowed to write during the first 15 minutes.
This time is to be spent in reading the question paper.
The time given at the head of this Paper is the time allowed for writing the answers.
Attempt seven questions in all.
Part I is compulsory. All questions from Part I are to be attempted.
A total of five questions are to be attempted from Part II.
The intended marks for questions or parts of questions are given in brackets [ ].
To be supplied with this Paper : Survey of India Map Sheet No. 45D/10
and 20 cm of twine.

Note:
(i) In all Map Work, make wise use of arrows to avoid overcrowding of the map.
(ii) The extract of Survey of India Map Sheet No. 45D/10 must not be taken out of the examination hall. It must be handed over to the Supervising Examiner on completion of the Paper.
(iii) The Map given at the end of this question paper must be detached, and after marking must be fastened to your answer booklet.
(iv) All sub-sections of the questions attempted must be answered in the correct serial
order.
(v) All working including rough work should be done on the same answer sheet which is used to answer the rest of the paper.

PART I [30 Marks]

Attempt all questions from this Part.

Question 1:
Study the extract of the Survey of India Map Sheet No. 45D/10 and answer the following questions :
(a) Give the four figure grid reference for a figure similar to the one given below. Identify the figure: [2]
ICSE Geography Question Paper 2015 Solved for Class 10 - 1(b) How is the drainage pattern in grid square 1606 different from that in grid square 1608? [2]

(c) Identify the correct six figure grid reference for each of the following:
(i) Gautam Maharishi Mandir
200071; 071200; 201070 ?
(ii) 0.443
172059; 052179; 179052 ? [2]

(d) Name the most prominent settlement other than ABU. Give two reasons to support your answer. [2]

(e) (i) What is the general slope of the land in the north-west comer of the map extract ?
(ii) What is the compass direction of Chandela (1803) from Hanumanji ka Mandir (2208) ? [2]

(f) What do you understand by the following terms as used on the map extract :
(i) Causeway (1702)
(ii) Falls 25m (2307). [2]

(g) (i) If you were to cycle at 10 km an hour, how much time would it take to cover the north-south distance depicted on this map extract ?
(ii) Calculate the area enclosed by Eastings 19 to 22 and Northings 04 to 09. [2]

(h) (i) Identify one natural feature in grid square 1610
(ii) Identify one man made feature in grid square 1903. [2]

(i) Give two probable reasons, other than dry water features, to indicate that the region depicted on the map extract receives seasonal rainfall. [2]

(j) Calculate, in metres, the difference in height between the highest point on the map extract and the contour height given in grid square 2402. [2]

Answer:
(a) The four fig grid reference for the figure is 1903. And the figure is a seasonal tank with an embankment.

(b) Drainage pattern of grid square 1606 is Radial and of grid square 1608 is Trellis.

(c) The correct six figure are
(i) Gautam Maharishi Mandir-200071.
(ii) 0.443-179052.

(d) Vajna settlement is another prominent settlement because
(1) It has a police Chauki.
(2) It is located on the metalled road therefore good transport is available.

(e) (i) General slope of the land in the North west comer is towards North west.
(ii) Compass direction of Chandela from Hanumanji Ka Mandir is South west.

(f) (i) Cause way : It is a raised road or path that enable us to cross a seasonal river. It does not work during rain. It is not a bridge.
(ii) Fall 25 m indicate the presence of a waterfall i.e., 25 metres high.

(g) (i) one hour.
(ii) Area = L × B = 3 × 5 = 15 km2

(h) (i) Natural feature in 1610 is broken ground and seasonal stream.
(ii) One man made feature in grid square 1903 is hut and embankment.

(i) Two reasons to indicate that the region depicted on the map receives seasonal rainfall are :

  1. Cart tracks are motorable in the dry season.
  2. Presence of causeway and broken ground.

(j) Difference between the highest point on the map and the contour height given in grid square 2402 is 1129 metres.

Question 2:
On the outline map of India provided :
(a) Mark and. name the Nilgiris. [1]
(b) Shade and label the Malwa Plateau. [1]
(c) Shade and label the Malabar Coastal Plains. [1]
(d) Mark and name the river Gomti. [1]
(e) Shade and name the Andaman Sea. [1]
(f) Mark and name Allahabad. [1]
(g) Mark with a single arrow and name the winds that bring winter rain to north-west India. [1]
(h) Mark and name Digboi. [1]
(i) Mark an area with laterite soil below the Tropic of Cancer. [1]
(j) Mark and name the Karakoram Pass. [1]

Answer:
ICSE Geography Question Paper 2015 Solved for Class 10 - 2

PART II [50 Marks]

Attempt any five questions from this Part.

Question 3:
(a) Explain two factors that affect the climate of India giving a suitable example for each. [2]

(b) State two differences between the rainfalls that occur from June to September and that from December to February in North India. [2]

(c) Give a geographic reason for each of the following :
(i) Kerala has the longest rainy season.
(ii) The Konkan coast experiences orographic rainfall.
(iii) The city of Kanpur in Uttar Pradesh has a higher range of temperature than that of Chennai in Tamil Nadu. [3]

(d) Study the climatic data given below and answer the questions that follow :

Month JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
Temp °C 21.0 21.9 24.3- 27.2 28.0 26.4 26.1 25.4 25.0 26.0 23.8 21.2
Rainfall Cm 5.1 2.8 1.2 1.7 3.9 4.6 8.4 11.4 11.9 31.6 34.5 14.8

(i) Identify the hottest month.
(ii) Calculate the annual rainfall.
(iii) Name the winds that bring the maximum rainfall to this city. [3]

Answer:
(a) Two factors that affect the climate in India are :

  1. Altitude: As the height increases temperature decreases at the rate of 1°C for every 165 m of ascend. Example : Nainital has lower temperature than Agra.
  2. Distance from the sea: Places close to the sea have lower range of temperature i.e. moderate climate and places away from the influence of the sea have higher range of temperature i.e. continental climate. Example : Mumbai has lower range of temperature than Delhi as it is close to the sea.

(b) Two differences between the rainfall that occur from June to September and from December to February are :

  1. Rainfall in June to September is caused by S.W. monsoon whereas rainfall in December to February is caused by temperate cyclones (western disturbances) originating in Mediterranean sea.
  2. Rain in June to September is heavy ranging between 75 to 150 cm. whereas rain in Dec. to February is only 5 to 7 cm but beneficial for wheat and barley.

(c) Geographical Reasons:
(i) Kerala has the longest rainy season because it lies in the extreme south of India and it is the first and last to see the monsoon.
(ii) Konkan coast experiences orographic rainfall because it lies of the wind ward side of western ghats when south west monsoon strikes it.
(iii) Kanpur in Uttar Pradesh has a higher range of temperature than that of Chennai because Kanpur is located away from the sea and the equator.

(d) (i) May is the hottest month.
(ii) Annual rainfall is 131.9 cm.
(iii) Maximum rainfall is received by North-East monsoon winds.

Question 4:
(a) State the characteristic of each of the soils named below that makes them most suitable for crop cultivation :
(i) Black soil.
(ii) Red soil. [2]

(b) State the geographic term for each of the following processes :
(i) The process by which soluble minerals dissolve in rain water and percolate to the bottom, leaving the top soil infertile.
(ii) The process by which rain water, flowing in definite paths, removes the top soil, thus causing deep cuts to the surface of the land. [2]

(c) Define the following :
(i) Pedogenesis.
(ii) Humus.
(iii) Bhangar. [3]

(d) Give a geographic reason for each of the following :
(i) Alluvial soil is extremely fertile.
(ii) Need for soil conservation.
(iii) Reafforestation should be practised extensively. [3]

Answer:
(a) (i) Characteristic of black soil : It is able to retain moisture.
(ii) Characteristic of red soil : It is rich in potash and become fertile with proper use of fertilizers and irrigation.

(b) Geographic terms are :
(i) Leaching
(ii) Gully erosion.

(c) Definition:
(i) Pedogenesis: Process of soil formation is called pedogenesis.
(ii) Humus: Decayed remains of plants, animal manures and dead animals is called Humus. It is an essential element in determining the fertility of soil.
(iii) Bhangar: Older alluvium soil found about 30 m above sea level in river terraces, light grey in colour and calcareous clay is called humus.

(d) Geographical reasons:
(i) Alluvial soil is extremely fertile because it is found to a depth of 500 m and rich in humus, lime and potash.
(ii) There is a need for soil conservation because top soil that is eroded is the main feeding zone. With the increase in population the demand of crops is also increasing.
(iii) Reafforestation should be practised extensively because the area under forest cover is shrinking day by day due to urbanization etc.

Question 5:
(a) State two characteristics of Tropical Deciduous forests. [2]

(b) State two reasons why Tropical Evergreen forests are difficult to exploit. [2]

(c) Identify the tree as per its characteristics mentioned below :
(i) It yields wood that is hard and scented and is usually found in high altitudes.
(ii) It is generally found in deltaic regions and is used to make boats.
(iii) The furniture made from the wood of this tree is generally the most expensive. [3]

(d) Differentiate between afforestation and deforestation. State a disadvantage of deforestation. [3]

Answer:
(a) Characteristics of Tropical deciduous forests are :
(i) They shed their leaves before the summer season.
(ii) These trees are found in pure stands.

(b) Tropical evergreen forests are difficult to exploit because they are found in mixed stands and found in dense growth.

(c) Trees as per its characteristics are :
(i) Deodar
(ii) Sundri
(iii) Teak.

(d) Afforestation means growing of trees where there are no trees. It increases the forested area.
Deforestation means cutting of trees for various purposes.
Disadvantage : It increases soil erosion.

Question 6:
(a) State two reasons why’irrigation is important to a country like India. [2]

(b) Name two modern methods of irrigation. State one important reason for their growing popularity. [2]

(c) (i) Why is well irrigation still a popular means of irrigation ? Give two reasons to support your answer.
(ii) State the significance of rainwater harvesting. [3]

(d) (i) Why is the world in danger of facing a severe water shortage in the coming future ? Give two reasons to support your answer.
(ii) State one measure the Government should adopt to handle the present water crisis. [3]

Answer:
(a) Irrigation is important to a country like India because rainfall here is seasonal in nature. It is limited to four months of a year. It is also important because some crops require more water than what it is provided by the rainfall therefore we have to depend on irrigation.

(b) Two modern methods of irrigation are :

  1. Perennial canals.
  2. Drip irrigation system.

They are becoming popular because there is minimum loss of water and can irrigate fields throughout the year.

(c) (i) Well irrigation is still a popular means of irrigation because :

  1. It can supply water whenever and where ever required.
  2. It is also a cheap source of irrigation.

(ii) Significance of rain water harvesting are :
(i) Reduce surface run off therefore no flooding of roads.
(ii) It raises ground water table by adding to ground water reserves.
(iii) It solve the problem of water scarcity.

(d) (i) The world is in danger of facing a severe water shortage in coming future because :

  1. Most of our surface waters are polluted by industrial waste, sewage etc.
  2. The indiscriminate use of chemicals, fertilizers, pesticides, insecticides, to increase farm productions had led to increasing ground water pollution.

(ii) One measure the Government should adopt to handle present water crisis is rain water harvesting and recycle and reuse of water.

Question 7:
(a) State two reasons why limestone is a valuable mineral. [2]

(b) State the most important use of the following :
(i) Iron ore
(ii) Bauxite [2]

(c) Name the :
(i) Largest oil refinery in the Public sector.
(ii) State that is the largest producer of coal.
(iii) Best variety of iron ore. [3]

(d) Give a geographic reason for each of the following :
(i) Many port cities have their own oil refineries.
(ii) Petroleum is called a ‘fossil fuel’.
(iii) Coal is called a versatile mineral. [3]

Answer:
(a) Limestone is a valuable mineral because it is used in various industries like chemicals, iron and steel, cement, fertilizers etc.

(b) Most important use of:
(i) Iron ore : In the production of steel.
(ii) Bauxite: In the production of aluminium.

(c) (i) Largest oil refinery in the public sector — Indian Oil Corporation Ltd.
(ii) Jharkhand is the largest coal producing state.
(iii) Hematite is the best variety of iron ore.

(d) Geographical reasons for the following are :
(i) Port cities have their own oil refineries because two thirds of the petroleum is imported. The crude oil is imported from Gulf countries and Malaysia.
(ii) Petroleum is called fossil fuel because it is derived from plant and animal . life buried in sedimentary rocks millions of years ago.
(iii) Coal is called a versatile mineral because it forms a basic raw material for the production of chemicals, dyes, fertilizers, paints, synthetics, explosive apart from source of energy.

Question 8:
(a) Differentiate between a Rabi crop and a Kharif crop. [2]

(b) State an important difference between the climatic requirements for growing cotton and jute. [2]

(c) Give the geographic term for each of the following :
(i) Cultivation of sugarcane from the root stock of the cane which has been cut.
(ii) The residue left behind after the crushing of oilseeds.
(iii) The process by which latex is converted into a thick, spongy mass by adding acetic acid or formic acid. [3]

(d) Give a geographical reason for each of the following :
(i) Tea is cultivated on hill slopes.
(ii) The yield per hectare of sugarcane is higher in the Southern states.
(iii) Pulses are important food crops. [3]

Answer:
(a) Rabi crop is sown during late November and harvested in March e.g. : wheat, mustard etc.
Kharif crop is sown in June and harvested in early November e.g. : rice, cotton etc.

(b) Climatic requirements for growing cotton :
Temperature: 20°C to 32°C and atleast 200 frost free days.
Rainfall: Ranging between 50 cm to 120 cm abundant sunshine is required during ripening period.
Climatic requirement for Jute :
Temperature: 21°C to 35°C a hot climate with high atmospheric humidity. Rainfall : Ranging between 150 cm to 200 cm.

(c) Geographical terms are :
(i) Ratoon crop
(ii) Oil cake
(iii) Coagulate.

(d) Geographical reasons are :
(i) Tea is cultivated on hill slopes because tea plant can not tolerate standing water on its roots.
(ii) Yield per hectare of sugarcane is higher in southern states because of the use of modem scientific agricultural methods and the factories are located close to the fields.
(iii) Pulses are important food crops because they are rich source of protein for vegetarian people.

Question 9:
(a) (i) Why is the cotton textile industry called an agro-based industry ?
(ii) Give an important reason for it being more widespread than the jute industry. [2]

(b) (i) State one important point of similarity between the woollen industry and the silk industry.
(ii) Name the state that produces the most woollen and silk products respectively. [2]

(c) (i) State two major problems faced by the sugar industry.
(ii) Name two by-products of the sugar industry. [3]

(d) (i) State one of the main problems of the silk industry.
(ii) Name two products of the jute industry, other than rope and gunny bags.
(iii) Why are synthetic fibres popular ? [3]

Answer:
(a) (i) Cotton textile is called an agro-based industry because for its raw material it depends on cotton which is an agricultural product.
(ii) It is more widespread than the jute industry because India has tropical climate where cotton is mainly used for clothing.

(b) (i) Main similarity between the woollen and the silk industry : Both of them derive their raw material from animals and both require skilled labour.
(ii) State that produces woollen products is Punjab and state that produces silk products is Karnataka.

(c) (i) Two problems faced by sugar industry are :
(1) It is the most soil exhausting crop therefore cost increases.
(2) Sugar mills are old-fashioned and far from the sugar farms.
(ii) Two by products of the sugar industry are :
(1) Bagasse
(2) Pressmud.

(d) (i) One main problem of the silk industry is that it has a limited market as it is very expensive.
(ii) Two products of the jute industry are upholstry and carpets.
(iii) Synthetic fibres are popular because they are cheap, durable and easy to maintain.

Question 10:
(a) (i) Why is the iron and steel industry called a basic industry ?
(ii) Define a mini steel plant. [2]

(b) With which large scale industry would you identify the following manufacturing centres :
(i) Kanpur
(ii) Rourkela
(iii) Pune
(iv) Mangalore. [2]

(c) (i) State two reasons for the growing importance in the status of petrochemical industries.
(ii) Name two products of the petroleum industry. [3]

(d) (i) State two conditions necessary for the setting up of a heavy engineering industry.
(ii) Name a ship building yard on the east coast and a centre for making electric locomotives. [3]

Answer:
(a) (i) Iron and steel industry is called basic industry because it supports fertilizers and cement industry. It is linked with the economic development of a country.
(ii) Mini steel plant is a unit that uses cheaply available scrap iron in the electric arc furnaces to make steel which is futher rolled and shaped into necessary products.

(b) Large scale industry that we identify with :
(i) Kanpur : Woollen industry.
(ii) Rourkela : Iron and Steel plant.
(iii) Pune : Electronics Industry.
(iv) Mangalore : Petrochemical Industry.

(c) (i) Two reasons for the growing importance is the status of petrochemical, industries are :

  1. It is durable, cheap, light weight and attractive.
  2. It has replaced wood, glass, metal, natural rubber etc.

(ii) Two products of the petroleum industry are :

  1. PVC pipes.
  2. Synthetic fibre.

(d) (i) Two conditions necessary for the setting up of a heavy engineering industry are :

  1. Availability of market.
  2. Availability of cheap labour.

(ii) Ship building yard on east coast is Vishakhapatnam and Chittaranjan for making electric locomotives.

Question 11:
(a) (i) Why is the Railways an important means of transport as compared to Airways ?
(ii) State one economic benefit of the Golden Quadrilateral Project. [2]

(b) (i) State one important difference between an expressway and a highway.
(ii) Name the first expressway constructed in the country.
(iii) State a reason why the Northern Rivers are more suitable for navigation than the Deccan Rivers. [3]

(c) (i) “Waste segregation is important”. Give a reason to support your answer.
(ii) Why is nuclear waste harmful ?
(iii) Explain briefly how as a student, you can help in the reduction of waste generation. [3]

(d) (i) What is understood by biodegradable waste ?
(ii) State one source of gaseous waste. [2]

Answer:
(a) (i) Railways is an important means of transport as compared to airways as it is cheap and has more carrying capacity of passengers and goods.
(ii) Economic benefit of the Golden Quadrilateral project is that it leads to general improvement of the area with more employment.

(b) (i) Expressway is a six lane highway where two and three wheelers are not allowed. A highway is a two lane road where all vehicles are allowed.
(ii) Mumbai-Pune expressway is the first expressway constructed in the country.
(iii) Northern Rivers are more suitable for navigation than the Deccan Rivers because they are perennial in nature and flow over gentle sloping plains.

(c) (i) Waste segregation is important because the method of treating waste depends on the nature of waste like solid waste, liquid waste, biodegradable and non-biodegradable waste.
(ii) Nuclear waste is more harmful because radioactive wastes remain active for a long time therefore it can enter in human body through food and water and can cause damage of tissues, blood cells and cancer.
(iii) As a student I can segregate the waste into biodegradable and non biodegradable waste of my house, reduce the water by incineration and recycling it.

(d) (i) Biodegradable waste is a type of waste which can be broken down in a reasonable amount of time, into its base compounds by micro organisms and other living things regardless of what those compounds may be.
(ii) Release of smoke, ashes and aerosols from chimneys of factories, increase CO2. Methane comes out from cattle shed. Swamps, coal mines etc. Volcanic erruptions gives out SO2.

ICSE Class 10 Geography Previous Years Question Papers