Solved 2017 Question Paper ICSE Class 10 Chemistry
Solutions for Chemistry, Class 10, ICSE
Section I 40 Marks
12 questionsFill in the blanks from the choices given in brackets:
(i) The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called .............. (electron affinity, ionisation potential, electronegativity)
(ii) The compound that does not have a lone pair of electrons is .............. (water, ammonia, carbon tetra chloride)
(iii) When a metallic oxide is dissolved in water, the solution formed has a high concentration of .............. ions. (H+, H3O+, OH-)
(iv) Potassium sulphite on reacting with hydrochloric acid releases .............. gas. (Cl2, SO2, H2S)
(v) The compound formed when ethene reacts with Hydrogen is .............. (CH4, C2H6, C3H8)
Answer:
(i) The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called Ionization Potential
(ii) The compound that does not have a lone pair of electrons is carbon tetrachloride.
(iii) When a metallic oxide is dissolved in water, the solution formed has a high concentration of OH- ions.
(iv) Potassium sulphite on reacting with hydrochloric acid releases SO2 gas.
(v) The compound formed when ethene reacts with Hydrogen is C2H6.
Answer:
Copper chloride.
Reason — Copper chloride forms a precipitate that is soluble in excess of ammonium hydroxide.
Answer:
Consists of molecules
Reason — In covalent compound, valence shell electrons are mutually shared by atom of each element to achieve a stable electronic configuration. As there is no transfer of electrons between the atoms so the atoms remain electrically neutral. Hence they don't form ions and covalent compounds consists of molecules.
Identify the substance underlined, in each of the following cases:
(i) Cation that does not form a precipitate with ammonium hydroxide but forms one with sodium hydroxide.
(ii) The electrolyte used for electroplating an article with silver.
(iii) The particles present in a liquid such as kerosene, that is a non electrolyte.
(iv) An organic compound containing -- COOH functional group.
(v) A solid formed by reaction of two gases, one of which is acidic and the other basic in nature.
Answer:
(i) Calcium [Ca2+] ions
(ii) sodium argentocyanide solution
(iii) Molecules only
(iv) Carboxylic acids
(v) Ammonium chloride.
Write a balanced chemical equation for each of the following:
(i) Action of cold and dilute Nitric acid on Copper.
(ii) Reaction of Ammonia with heated copper oxide.
(iii) Preparation of methane from iodomethane.
(iv) Action of concentrated sulphuric acid on Sulphur.
(v) Laboratory preparation of ammonia from ammonium chloride.
Answer:
(i) 3Cu + 8HNO3 [dil.] ⟶ 3Cu(NO3)2 + 4H2O + 2NO [g]
(ii) 2NH3 + 3CuO ⟶ 3Cu + 3H2O + N2 [g]
(iii)
(iv) S + 2H2SO4 (conc.) ⟶ 3SO2 + 2H2O
(v) 2NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3
State one relevant observation for each of the following reactions:
(i) Addition of ethyl alcohol to acetic acid in the presence of concentrated Sulphuric acid.
(ii) Action of dilute Hydrochloric acid on iron (II) sulphide.
(iii) Action of Sodium hydroxide solution on ferrous sulphate solution.
(iv) Burning of ammonia in air.
(v) Action of concentrated Sulphuric acid on hydrated copper sulphate.
Answer:
(i) On warming, the mixture gives fruity smell. Acetic acid on heating with an alcohol and a dehydrating agent [conc. H2SO4] forms an ester — ethyl acetate.
(ii) H2S gas is evolved. It has a foul smell like rotten eggs.
FeS + 2HCl [g] ⟶ FeCl2 + H2S
(iii) A dirty green precipitate is formed which is insoluble in excess of sodium hydroxide solution.
FeSO4 + 2NaOH ⟶ Fe(OH)2↓ (dirty green ppt.) + Na2SO4
(iv) Ammonia burns in the atmosphere of excess oxygen with a green or greenish yellow flame, forming nitrogen and water vapour.
4NH3 + 3O2 ⟶ 2N2 + 6H2O
(v) The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.
(i) Draw the structural formula for each of the following:
- 2, 3 – dimethyl butane
- Diethyl ether
- Propanoic acid
(ii) From the list of terms given, choose the most appropriate term to match the given description.
(calcination, roasting, pulverisation, smelting)
- Crushing of the ore into a fine powder.
- Heating of the ore in the absence of air to a high temperature.
Answer:
1. 2, 3 – dimethyl butane :

2. Diethyl ether :

3. Propanoic acid :

(ii) The matching terms are:
- Crushing of the ore into a fine powder — Pulverisation
- Heating of the ore in the absence of air to a high temperature — Calcination
(i) Calculate the number of gram atoms in 4.6 grams of sodium (Na = 23). [5]
(ii) Calculate the percentage of water of crystallization in CuSO4.5H2O (H = 1, O = 16, S = 32, Cu = 64)
(iii) A compound of X and Y has the empirical formula XY2. Its vapour density is equal to its empirical formula weight. Determine its molecular formula.
Answer:
(i) 23 g of Na = 1 g. atom
∴ 4.6 g. of Na = x 4.6 = 0.2 g
Hence, 0.2 g. atoms
(ii) Molecular weight of hydrated copper sulphate CuSO4.5H2O = 64 + 32 + 4(16) + 5[2(1) + 16]
= 64 + 32 + 64 + 5[18]
= 64 + 32 + 64 + 90
= 250 g
250 g of CuSO4.5H2O contains 90 g of water of crystallization.
∴ Percentage of water of crystallization = x 100 = 36%
Hence, percentage of water of crystallization is 36%
(iii) Empirical formula = XY2
Empirical formula weight = V.D.
Molecular weight = 2 x V.D.
∴ Molecular formula = n[E.F.] = 2[XY2] = X2Y4
Answer:
(i) 15
(ii) 19
(iii) 8
(iv) 4
(v) 2
Section Ii 40 Marks
19 questionsAnswer:
(i) He < Ne < Ar
Reason — Number of electron shells increases as we move down the group.
(ii) K < Na < Li
Reason — Ionization Potential decreases as we move down the group.
(iii) Br < Cl < F
Reason — Electronegativity decreases as we move down the group.
(iv) Li < Na < K
Reason — Atomic Size increases as we move down the group.
Answer:
(i) Covalent bonding
Reason — The valency of hydrogen element is 1 and that of oxygen is 2. Hydrogen needs one electrons to attain stable duplet structure of nearest noble gas - He [2] and oxygen needs two electrons to attain stable octet structure of nearest noble gas - Ne [2,8]
Each of the two hydrogen atoms shares an electron pair with the oxygen atom such that hydrogen acquires a duplet configuration and oxygen an octet configuration resulting in the formation of two single covalent bonds [H-O-H] in the molecule of water.

(ii) Electrovalent bonding
Reason — The valency of calcium (2,8,8,2) element and that of oxygen (2,6) is 2. Calcium donates two electrons and oxygen takes up those two electrons and electrovalent bond is formed.
Answer:
(i) With Sodium hydroxide, copper sulphate (CuSO4) solution forms a pale blue precipitate which is insoluble in excess of sodium hydroxide and with ammonium hydroxide it forms a pale blue precipitate which dissolves in excess of ammonium hydroxide and forms a deep/inky blue solution.
(ii) Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.
H2SO4 (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed which does not dissolve on warming the mixture].
Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.
2HCl (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbCl2 ↓ [white ppt. formed which dissolves on warming the mixture].
Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.
Identify the salts P and Q from the observations given below:
(i) On performing the flame test salt P produces a lilac coloured flame and its solution gives a white precipitate with silver nitrate solution, which is soluble in Ammonium hydroxide solution.
(ii) When dilute HCl is added to a salt Q, a brisk effervescence is produced and the gas turns lime water milky. When NH4OH solution is added to the above mixture (after adding dilute HCl), it produces a white precipitate which is soluble in excess NH4OH solution.
Answer:
(i) P is potassium chloride.
Reason — K+ ions produces lilac coloured flame and Cl- ions react with silver nitrate to form silver chloride ppt. (soluble in excess of ammonium hydroxide).
(ii) Q is zinc carbonate.
Reason — CO32- ions produces carbon dioxide with HCl. Zinc chloride forms white ppt. with ammonium hydroxide ( soluble in excess of ammonium hydroxide).
Answer:
(i) Electron dot diagram showing the formation of Methane is given below:

(ii) Electron dot diagram showing the formation of Magnesium chloride is given below:

Answer:
(i) The observations at the anode & at the cathode are:
At cathode — silvery grey deposit of lead metal.
At anode — reddish brown fumes of bromine vapours.
(ii) The observations at the anode & at the cathode are:
At cathode — Brownish pink copper metal is deposited at cathode during electrolysis of copper sulphate.
At anode — Copper ions are formed. Copper anode diminishes in mass.
Blue colour of CuSO4 remains unchanged.
Answer:
(i) OH- ion will get discharged in preference to SO42- or NO3- ions.
(ii) Ag+ ions will get discharged in preference to Pb2+ or Cu2+ ions.
Answer:
A : Sodium hydrogen sulphate (NaHSO4) and HCl gas.
B : Upward displacement of air.
C : Magnesium nitride (Mg3N2) and water (H2O).
D : Quick lime [CaO]
E : Downward displacement of air.
Answer:
(i) C + 2H2SO4 (conc.) ⟶ CO2 + 2SO2 + 2H2O
(ii) Mg + H2SO4 (dil.) ⟶ MgSO4 + H2 (g)
(iii)
(i) Propane burns in air according to the following equation :
C3H8 + 5O2 ⟶ 3CO2 + 4H2O.
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen.
(ii) The mass of 11.2 litres of a certain gas at s.t.p. is 24 g. Find the gram molecular mass of the gas.
Answer:
(i) Given, 20% of air contains oxygen
∴ 20% of 1000 cm3 = x 1000 = 200 cm3
[By Lussac's law]
To calculate the volume of propane consumed :
Hence, volume of propane is consumed is 40 cm3
(ii) 11.2 lit. weighs 24 g
∴ 22.4 lit. will weigh = x 22.4 = 48 g
Hence, gram molecular mass of the gas = 48 g.
A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure:
(i) Find the number of moles of hydrogen present.
(ii) What weight of CO2 can the cylinder hold under similar conditions of temperature and pressure? (H= 1, C = 12, O = 16)
(iii) If the number of molecules of hydrogen in the cylinder is X, calculate the number of CO2 molecules in the cylinder under the same conditions of temperature and pressure.
(iv) State the law that helped you to arrive at the above result.
Answer:
(i) 2 g of hydrogen gas = 1 mole
∴ 1000 g of hydrogen gas = = 500 moles.
(ii) Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g
1 mole of carbon dioxide = 44 g
∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.
Hence, Weight of carbon dioxide in cylinder = 22 kg
(iii) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
∴ molecules in the cylinder of carbon dioxide = X.
(iv) Avogadro's law — Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.
Answer:
(i) S + 6HNO3 [conc.] ⟶ H2SO4 + 2H2O + 6NO2
(ii)
(iii) KNO3 + H2SO4 [conc.] KHSO4 + HNO3
(iv) NH3 + HNO3 ⟶ NH4NO3
Identify the term or substance based on the descriptions given below:
(i) Ice like crystals formed on cooling an organic acid sufficiently.
(ii) Hydrocarbon containing a triple bond used for welding purposes.
(iii) The property by virtue of which the compound has the same molecular formula but different structural formulae.
(iv) The compound formed where two alkyl groups are linked by group.
Answer:
(i) Glacial acetic acid
(ii) Ethyne or acetylene
(iii) Isomerism
(iv) Ketone or Alkanone
Answer:
(i) Galvanization
(ii) Solder [fuse metal]
(iii) Zinc blende [ZnS]
(iv) Copper oxide [CuO]
Answer the following questions with respect to the electrolytic process in the extraction of aluminum:
(i) Identify the components of the electrolyte other than pure alumina and the role played by each.
(ii) Explain why powdered coke is sprinkled over the electrolytic mixture.
Answer:
(i) Cryolite and fluorspar are components of the electrolyte. Addition of cryolite enhances the conductivity of the mixture since, pure alumina is almost a non-conductor of electricity. Addition of both cryolite & fluorspar lowers the fusion point of the mixture i.e., mixture fuses around 950°C instead of 2050°C.
(ii) The layer of powdered coke is sprinkled over the electrolytic mixture as :
- it prevents burning of carbon electrodes in air at the emergence point from the bath.
- it minimizes or prevents heat loss by radiation.
Complete the following by selecting the correct option from the choices given:
(i) The metal which does not react with water or dilute H2SO4 but reacts with concentrated H2SO4 is ............... (Al/Cu/Zn/Fe)
(ii) The metal whose oxide, which is amphoteric, is reduced to metal by carbon reduction ............... (Fe/Mg/Pb/Al)
(iii) The divalent metal whose oxide is reduced to metal by electrolysis of its fused salt is ............... (Al/Na/Mg/K)
Answer:
(i) The metal which does not react with water or dilute H2SO4 but reacts with concentrated H2SO4 is Cu.
(ii) The metal whose oxide, which is amphoteric, is reduced to metal by carbon reduction — Pb.
(iii) The divalent metal whose oxide is reduced to metal by electrolysis of it's fused salt is — Mg.