ICSE Class 9 Chemistry Sample Question Paper 10 with Answers
Section -1
(Attempt all questions from this section)
Question 1.
(a) Fill in the blanks with the correct choice given in brackets.
(i) ………….. . does not support combustion. [hydrogen/oxygen]
(ii) A reducing agent is an………………… of electrons. [acceptor/donor]
(iii) The reaction of photosynthesis is an…………… reaction, [endothermic/exothermic]
(iv) …………. water forms lather with soap. [Hard /Soft]
(v) The…………… temperature is called absolute zero. [+ 273°C/- 273°C]
Answer:
(i) Hydrogen
(ii) Donor
(iii) Endothermic
(iv) Soft
(v) – 273°C
(b) Choose the correct answer from the options given below :
(i) (Choose the air pollutant which is non-acidic.
(A) NO2
(B) SO2
(C) SO3
(D) Ozone
Answer:
(D) Ozone ‘
(ii) Choose the odd one.
(A) HCl
(B) H2CO3
(C) HNO3
(D) H2SO4
Answer:
(A) HCl
(iii) On adding water to sodium, the solution formed is :
(A) Neutral
(B) Alkaline
(C) Acidic
(D) Amphoteric
Answer:
(B) Alkaline
(iv) According to Boyle’s law, as the pressure increases, the volume :
(A) Increases
(B) Decreases
(C) Remains the same
(D) First increases and then decreases
Answer:
(B) Decreases
(v) In the element 11Na23,11 represents :
(A) Mass number
(B) Atomic number
(C) Number of neutrons
(D) None of the above
Answer:
(B) Atomic number
(c) Give the valency and the formula of the following radicals :
(i) Sulphate
(ii) Sulphite
(iii) Sulphide
(iv) Carbonate
(v) Ammonium
Answer:
| Formula | Valency | |
| (i) | SO42- | -2 |
| (ii) | SO32- | -2 |
| (iii) | S2- | -2 |
| (iv) | CO32- | -2 |
| (v) | NH4- | +1 |
(d) An element ‘M’ has three electrons more than the noble gas. Give the formula of its.
(i) Chloride
(ii) Sulphate
(iii) Hydroxide
(iv) Phosphate
(v) Oxide
Answer:
The outermost shell of all the noble gases is complete. Thus, its valency is zero. ‘M’ has three electrons more than the noble gas. Thus, the valency of the element ‘M’ is +3.
(i) MCl3
(ii) M2(SO4)3
(iii) M(OH)3
(iv) MPO4
(v) M2O3 ‘
(e) Correct the following statements.
(i) A molecular formula represents an element.
(ii) The molecular formula of water (H2O) represents nine parts by mass of water.
(iv) A balanced equation obeys the law of conservation of mass and so does an imbalanced equation.
(v) A molecule of an element is always monoatomic.
(vi) CO and Co both represent cobalt.
Answer:
(i) A molecular formula represents a molecule.
(ii) Molecular formula of water (H2O) represents eighteen parts by mass of water.
(iii) A balanced equation obeys the law of conservation of mass, while an unbalanced equation does not obey this law.
(iv) A molecule of an element is not always monoatomic.
(v) CO and Co represent carbon monoxide and cobalt, respectively.
(f) (i) What is a solubility curve ?
(ii) Write down two applications of solubility curve.
(iii) 12 g of a saturated solution of potassium chloride at 20°C, when evaporated to dryness, leaves a solid residue of 3 g. Calculate the solubility of potassium chloride.
Answer:
(i) A solubility curve is a line graph that plots changes in solubility of a solute in a given solvent against changing temperature.
(ii) Applications of solubility curve :
1. Shape of the curve indicates how the solubility of the given substance in a solvent varies with change in temperature. The solubility of a substance at a particular temperature can be determined from the curve.
2. The effect of cooling of hot solutions of different substances can be found from the curves.
(iii) Weight of water in solution = 12g – 3g = 9g
9g of water dissolves 3g of solid
Therefore 100 g of water will dissolve = \(\frac{3}{9} x 100 = 33.3 g\)
Solubility of KCl in water at 20°C is 33.3 g.
(g) Give reasons for the following.
(i) Electrovalent compounds conduct electricity in molten or aqueous state.
(ii) Electrovalent compounds have high melting and boiling points, while covalent compounds have low melting and boiling points.
(iii) 1. Electrovalent compounds dissolve in water, whereas covalent compounds do not.
2. Polar covalent compounds conduct electricity.
3. Electrovalent compounds are usually hard crystals yet brittle.
Answer:
(i) 1. They are good conductors of electricity in the fused or aqueous state because electrostatic forces of attraction between ions in the solid state are very strong, and these forces weaken in the fused state or in the solution state. Hence, ions become mobile.
2. In electrovaient compoimds, there exists a strong force of attraction between the oppositely charged ions, and a large amount of energy is required to break the strong bonding force between ions. So, they have high boiling and melting points. In covalent compounds, weak forces of attraction exist between the binding molecules, thus less energy is required to break the force of binding. So, they have low boiling and melting points.
(ii) 1. As water is a polar compound, it decreases the electrostatic forces of attraction, resulting in free ions in aqueous solution. Hence, electrovalent compounds dissolve. Covalent compounds do not dissolve in water but dissolve in organic solvents. Organic solvents are non-polar; hence, these dissolve in non-polar covalent compounds.
2. Polar covalent compounds conduct electricity because they form ions in their solutions.
3. Electrovaient compounds are usually hard crystals yet brittle because they have strong electrostatic forces of attraction between their ions which cannot be separated easily.
(h) Match Column A with Column B.
| Column A | Column B | |
| (i) | Plaster of Paris | (A) Electrovalent compound |
| (ii) | Nad | (B) Polar compound |
| (ii) | Nad | (B) Polar compound |
| (v) | Hydrogen chloride | (E) Non polar compound |
Answer:
(i) (D)
(ii) (A)
(iii) (E)
(iv) (B)
(v) (C)
Section – II
(Attempt any four questions from this section)
Question 2.
(a) Explain the Rutherford’s alpha particles scattering experiment with the help of a diagram.
(b) How temporary hardness’s removed by Clark’s process and also explain why removal of temporary hardness by boiling water is not a practical method ?
(c) According to the activity series, which of the following can successfully displace hydrogen ? K/Na/Pb/Ag/Pt/Fe/Al.
Answer:
(a) In 1911, Earnest Rutherford, a scientist from New Zealand, overturned Thomson’s atomic model by his gold foil experiment. His experiment demonstrated that the atom has a tiny massive nucleus. It thus rejected Thomson’s model of the atom.
Rutherford’s scattering experiment
1. Rutherford selected a gold foil as he wanted a very thin layer.
2. In his experiment, fast-moving alpha particles were made to fall on a thin gold foil.
3. Alpha particles are helium ions with +2 charges and have a considerable amount of energy.
4. These particles were studied by the flashes of light they produced on striking a zinc sulphide screen.
5. He expected the alpha particles to pass through the gold foil with little deflections and to strike the fluorescent screen.
Main features of Rutherford’s theory of atom are :
1. There is a positively charged centre in the atom called the nucleus in which nearly all the mass of the atom is concentrated.
2. Negatively charged particles called electrons revolve around the nucleus in paths called orbits.
3. The size of the nucleus is very small as compared to the size of the atom.
4. His model can be compared to the solar system, where the planets are compared with electrons and the sun with the nucleus.
(b) Removal of temporary hardness by Clark’s process is also known by the name by addition of lime. In this method lime is first thoroughly mixed with water in a tank and fed into another tank containing the hard water. Revolving paddles thoroughly mix the two solutions. Most of the calcium carbonate settles down. If there is any solid left over, it is removed by a filter. In this process the following reactions takes place.
Ca(HCO3)2 + Ca(OH)2 → 2CaCO3 + 2H2O
Mg(HCO3)2 + Ca(OH)2 MgCO3 + CaCO3 + 2H2O
Temporary hardness is called temporary because the temporary hardness of water can be removed just by boiling. But it is not a practical method to do, because it is slow and very costly.
(c) K and Na can displace hydrogen from acids by reacting violently.
Pb displaces hydrogen from only hot concentrated acids.
Ag and Pt do not displace hydrogen from acids at all.
Fe displaces hydrogen gently from acids.
A1 displaces hydrogen from acids vigorously.
Question 3.
(a) Write a note on discovery of protons and also write two properties of anode ray.
Answer:
(a) Goldstein scientist noticed another set of rays travelling in a direction opposite to that of cathode rays. These rays travel from anode to cathode. He called these rays as canal rays since these rays passed through holes or “canals” in the cathode. These rays were named as positive rays or anode rays. Anode rays travel in a straight line and are deflected by electric and magnetic fields but in a direction opposite to that of the cathode rays. This shows that these rays consist of positively charged particles called protons.
Properties of anode rays :
1. Anode rays consist of minute material particles and hence produce mechanical effects.
2. These rays produce fluorescence on a zinc sulphide screen.
(b) Under what conditions can hydrogen be made to combine with ?
(i) Nitrogen
(ii) Chlorine
(iii) Sulphur
(iv) Oxygen
Name the products in each case and write the equation for each reaction.
Answer:
(i) Three volumes of hydrogen and one volume of nitrogen react at temperature 450- 500°C and pressure 200-900 atm in the presence of a finely divided iron catalyst with molybdenum as promoter to give ammonia.
\(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\)
(ii) Equal volumes of hydrogen and chlorine react slowly in diffused sunlight to form hydrogen chloride.
H2 + Cl2 → 2HCl
(iii) Hydrogen gas on passing through molten sulphur reacts to give hydrogen sulphides.
H2 + S → H2S
(iv) Hydrogen bums in the presence of electric spark with a ‘pop’ sound in oxygen and with a blue flame forming water.
2H2 + O2 → 2H2O
(c) The formula of the chloride of a metal ‘M’ is MCl2. State the formula of its :
(i) Carbonate
(ii) Nitrate
(iii) Hydroxide.
Answer:
(i) MCO
(ii) M(NOs)2
(iii) M(OH)2
Question 4.
(a) Two neutral gases ‘A’ and ‘B’ undergo a synthesis reaction to form a gas ‘C’.
(i) Identify ‘A’, ‘B’ and ‘C’.
(ii) Name the process by which gas ‘C’ is manufactured. Give the balanced chemical equation along with the conditions.
(iii) What do you observe when gas ‘C comes in contact with ?
(1) Moist red litmus paper,
(2) Concentrated hydrochloric acid.
Answer:
(a) (i) A – Nitrogen B – Hydrogen C – Ammonia
(ii) Gas ‘C’ is manufactured by the Haber process.
N2 + 3H2 → 2NH3 + Heat
Favourable conditions :
- Temperature should be between 450°C and 500°C.
- Pressure should be high (200-1000 atm.).
- Promoter used should be molybdenum.
(iii) It turns moist red litmus blue.
(b) What do you understand by the combining capacity of atoms? Explain with examples.
Answer:
An atom of each element has a definite combining capacity called its valency. The combining capacity of the atoms, i.e., their tendency to react and form molecules with atoms of the same or different elements, was explained as an attempt to attain a fully filled outermost shell of electrons.
This is done by sharing, gaining or losing electrons. For example, lithium and sodium atoms contain one electron each in their outermost shell; therefore, each of them can lose one electron to have 8 electrons in their outermost shell. So, they are said to have valency = 1.
(c) What is meant by :
(i) electronic configuration,
(ii) atomic number
(iii) atomic mass number
Answer:
(i) Electronic configuration is the arrangement of electrons in the atomic or molecular orbitals of atoms or molecules.
(ii) Atomic number of elements is the number of protons in an atom of the element.
(iii) Atomic mass of an element is the total mass of electrons, protons and neutrons in one atom of the element.
Question 5.
(a) What is the effect of increasing and decreasing pressure on the solubility of a gas in a liquid ?
Answer:
On increasing pressure the solubility of a gas in a liquid increases whereas, on decreasing pressure the solubility of a gas in a liquid decreases. This shows the mass of a given volume of a gas which dissolves in liquid at constant temperature is directly proportional to the pressure on the surface of the liquid and thus in accordance with Henry’s law.
(b) State which salts increase in weight, decrease in weight or remain the same when exposed to the atmosphere.
(i) Sodium hydroxide
(ii) Ferric chloride
(iii) Green vitriol
(iv) Cone, sulphuric acid
Answer:
(i) Increases
(ii) Increases
(iii) Decreases
(iv) Increases
(c) Why is it necessary to compare gases at STP ?
Answer:
The volume of a given mass of dry enclosed gas depends on the pressure of the gas and the temperature of the gas in Kelvin, so to express the volume of the gases, we compare these to STP.
Question 6.
(a) How does the modern atomic theory contradict and correlate with Dalton’s atomic theory ?
Answer:
Dalton was right that atoms take part in chemical reactions.
Comparisons of Dalton’s atomic theory with the modem atomic theory.
Dalton’s atomic theory :
1. Atoms are indivisible.
2. Atoms of the same element are similar in every respect.
3. Atoms combine in a simple whole number ratio to form molecules.
4. Atoms of different elements are different.
5. Atoms can neither be created nor be destroyed.
Modem atomic theory :
1. Atoms are no longer indivisible and consist of electrons, protons, neutrons and even more sub-particles.
2. Atoms of the same element may differ from one another called isotopes.
3. Atoms of different elements may be similar called isobars.
4. Atoms combine in a ratio which is not a simple whole number ratio; e.g., in sugar, the C12H22O11 ratio is not a whole number ratio.
(b) Classify the following as homogeneous or heterogeneous and give one example of each :
(i) Solid-Solid
(ii) Solid-Liquid
(iii) Gas-Gas
(iv) Liquid-Liquid
(v) Gas-Solid
Answer:
| Constituents of mixture | Nature of mixture | Examples |
| (i) Solid-Solid | Homogeneous | Alloys |
| (ii) Solid-Liquid | Homogeneous | Salt in water |
| (iii) Gas-Gas | Homogeneous | Air |
| (iv) Liquid-Liquid | Homogeneous | Milk in water |
| (v) Gas-Solid | Heterogeneous | Smoke |
(c) Draw the orbit structure for each of the following compounds :
(i) Methane [H = 1, C = 6]
(ii) Magnesium chloride [Mg = 12, Cl = 17]
Answer:
(i) Orbit structure of methane
(ii) Orbit structure of magnesium chloride
The two electrons lost by a magnesium atom are gained by chlorine atoms to produce a magnesium ion and two chloride ions.
Question 7.
(a) Give reasons for the following :
(i) Hydrogen shows dual nature
(ii) Gases diffuse rapidly.
(iii) Rivers and lakes do not freeze easily ?
Answer:
(i) Hydrogen shows dual nature because it resembles the alkali metals of group LA and the halogens of group VELA.
(ii) Gases have maximum intermolecular space. Therefore, when two gases are brought in contact, they readily fill the intermolecular spaces and form a homogeneous mixture.
(iii) Rivers and lakes have a large amount of water, and water has high specific heat capacity due to which it does not freeze easily. Even if water freezes, it freezes into ice on the surface (at the top) of rivers and lakes. Water is present below because of anomalous expansion of water.
(b) Name the solvent for the following precipitates :
(i) Silver chloride
(ii) Lead sulphate
(iii) Lead chloride
Answer:
(i) Ammonium hydroxide
(ii) Ammonium acetate
(iii) Soluble in hot water
(c) What is latent heat of vapourization ?
Answer:
Latent heat of vapourization is the heat energy required to change 1 kg of a liquid to a gas at atmospheric pressure at its boiling point.