Solved 2022 Semester 2 Question Paper ICSE Class 10 Chemistry
Solutions for Chemistry, Class 10, ICSE
Section A 10 Marks
10 questionsAnswer:
It is highly soluble in water
Reason — Hydrogen chloride gas is not collected over water since it is highly soluble in water.
Answer:
Fluorspar
Reason — Fluorspar lowers the fusion point of the mixture in Hall Heroult Process and the mixture fuses at 950° C instead of 2050° C.
Section B 30 Marks
20 questionsAnswer:
(a) Isomerism — Isomerism is the phenomenon due to which two or more compounds have the same molecular formula but differ in molecular arrangement or in structural formula.
(b) Ores — Ores are minerals from which metals are extracted commercially at a comparatively lower cost and with minimum effort.
Answer:
(a) Carbon dioxide [CO2]
(b) Nitrogen dioxide [NO2]
Answer:
(a) Dry ammonia gas reacts with oxygen in the presence of platinum (catalyst) to produce nitric oxide (NO), water and heat. Nitric oxide (NO) produced is oxidised to nitrogen dioxide (NO2) and reddish brown vapours are seen in the flask. As the reaction is exothermic, so the platinum (catalyst) continues to glow even after the heating is discontinued.
4NH3 + 5O2 6H2O + 4NO↑ + Δ
2NO + O2 ⟶ 2NO2
(b) Colourless ammonia gas reacts with greenish yellow chlorine gas in excess producing hydrogen chloride and a yellow explosive liquid (nitrogen trichloride).
NH3 + 3Cl2 [excess] ⟶ 3HCl + NCl3
(c) Carbon reacts with hot concentrated nitric acid producing carbon dioxide, water and reddish brown fumes of nitrogen dioxide are evolved.
C + 4HNO3 ⟶ CO2 + 2H2O + 4NO2
Answer:
(a) C12H22O11 12C (sugar charcoal) + 11H2O
(b) Fe(OH)3 + 3HNO3 [dil.] ⟶ Fe(NO3)3 + 3H2O
(c) 2NH4OH + H2SO4 ⟶ (NH4)2SO4 + 2H2O
Answer:
(a) Bauxite is heated under pressure with conc. caustic soda (NaOH) for 2 to 8 hours. Bauxite dissolves and forms sodium meta aluminate because of the amphoteric nature of aluminium, leaving behind insoluble impurities called red mud. Red mud consists of ferric oxide, sand, etc. which are removed by filtration.
(b) Aluminium oxide due to its great affinity for oxygen is a very stable compound. It is not reduced easily by common reducing agents like carbon, carbon monoxide and hydrogen. Hence, fused alumina can be reduced to aluminium only by electrolysis.
Complete the table given below which refers to the Laboratory preparation of Ammonia gas:
Laboratory preparation | Reactants used | Products formed | Drying agent | Method of collection |
---|---|---|---|---|
Ammonia gas | (a) ............... | Calcium chloride + water + ammonia | (b) ............... | (c) ............... |
Answer:
Laboratory preparation | Reactants used | Products formed | Drying agent | Method of collection |
---|---|---|---|---|
Ammonia gas | (a) ammonium chloride [NH4Cl] and calcium hydroxide [Ca(OH)2)] | Calcium chloride + water + ammonia | (b) quicklime [CaO] | (c) downward displacement of air |
Answer:
(a) Hall-Heroult's process
(b) Fountain experiment
(c) Dibasic property of sulphuric acid.
Answer:
(a) MnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2↑
(b) Zn + 2HCl ⟶ ZnCl2 + H2↑
Answer:
In electrolysis of fused Alumina, the anode is made of (a) graphite and the product formed at cathode is (b) aluminium
Answer:
(a) Nitrate ion [NO31-]
(b) As iron [II] sulphate on exposure to the atmosphere is oxidised to iron [III] sulphate and the test will not answer with iron [III] sulphate, hence, freshly prepared iron[II] sulphate used in the test.
(c) Concentrated Sulphuric Acid.
Answer:
(a) On adding dilute sulphuric acid to sodium sulphite solution and warming, a colourless gas with suffocating smell of burning sulphur is evolved. The evolved gas turns lime water milky and changes the colour of acidified K2Cr2O7 solution from orange to green suggesting that it is SO2.
Na2SO3 + H2SO4 (dil.) ⟶ Na2SO4 + H2O + SO2↑
Sodium sulphate does not react with dilute sulphuric acid as an acid does not react with its own salt due to the presence of same anion.
(b) A lead salt gives a chalky white ppt. on reaction with ammonium hydroxide that is insoluble in excess of ammonium hydroxide. For example:
Pb(NO3)2 + 2NH4OH ⟶ 2NH4NO3 + Pb(OH)2 ↓
On the other hand, zinc salt forms a white gelatinous ppt. which dissolves when excess of ammonium hydroxide is added. Hence, the two can be distinguished.
ZnSO4 + 2NH4OH ⟶ (NH4)2SO4 + Zn(OH)2 ↓
Zn(OH)2 + (NH4)2SO4 + 2NH4OH ⟶ [Zn(NH3)4]SO4 + 4H2O
Identify the acid in each case:
(a) The acid formed when sulphur reacts with concentrated nitric acid.
(b) An acid, which on adding to lead nitrate solution produces a white precipitate which is soluble on heating.
(c) The acid formed when potassium nitrate reacts with a least volatile acid.
Answer:
(a) Sulphuric acid.
(b) Hydrochloric acid
(c) Nitric acid