Previous Year Paper year:2019 : ICSE-X-Chemistry

Home Learn Fast select Course

Change Unit Change Subject Last Menu

ICSE-X-Chemistry

Previous Year Paper year:2019

with Solutions - page 2

#1-c
Write a balanced chemical equation for each of the following reactions:
[5]

#1-c-i
Reduction of copper (II) oxide by hydrogen.

Ans : CuO+H₂→ Cu+H₂O

#1-c-ii
Action of dilute sulphuric acid on sodium hydroxide.

Ans : 2NaOH + H₂S0₄ → Na₂SO₄ + 2H₂O

#1-c-iii
Action of dilute sulphuric acid on zinc sulphide.

Ans : ZnS + H₂SO₄ → ZnSO₄ + H₂S

#1-c-iv
Ammonium hydroxide is added to ferrous sulphate solution.

Ans : FeSO₄ + 2NH₄0H → (NH₄)SO₄ + Fe(OH)₂↓

#1-c-v
Chlorine gas is reacted with ethene.

Ans :

#1-d
State one observation for each of the following:
[5]

#1-d-i
Concentrated nitric acid is reacted with sulphur.

Ans :

Dense brown fumes of nitrogen dioxide gas will be released

#1-d-ii
Ammonia gas is passed over heated copper (II) oxide.

Ans :

The black colour copper oxide will change to reddish pink copper.

#1-d-iii
Copper sulphate solution is electrolysed using copper electrodes.

Ans : When copper sulphate solution is electrolysed using copper electrodes, reddish pink deposit of copper metal takes place on cathode.

#1-d-iv
A small piece of zinc is added to dilute hydrochloric acid.

Ans : Zn + 2HCl > ZnCl₂ + H₂↑

Zinc metal dissolves forming solution with the liberation of hydrogen gas which bums with blue flame and gets extinguished with pop sound.

#1-d-v
Lead nitrate is heated strongly in a test tube.

Ans :

White powder of lead nitrate decompose to form yellow coloured lead oxide and liberating dense brown fumes of nitrogen dioxide gas.

#1-e

#1-e-i
Calculate:
[5]

1. The number of moles in 12 g of oxygen gas. [O = 16]

2. The weight of 1022 atoms of carbon.

[C = 12, Avogadro’s No. = 6 × 10²³]

Ans : 1. Oxygen gas (O₂)

Molecular mass = 16 x 2 = 32 g

32 g of oxygen gas → 1 mole

1 g of oxygen gas → $\frac { 1 }{ 32 }$ ``\frac { 1 }{ 32 }`` mole

12 g of oxygen gas → $\frac { 1 }{ 32 }$ ``\frac { 1 }{ 32 }`` × 12mole = 0.375 mole

2. 6.022 x 10²³ atoms of carbon weigh → 12 g

1 atom of carbon weighs → $\frac{12}{6 \times 10^{23}}$ ``\frac{12}{6 \times 10^{23}}``

10²² atoms of carbon will weigh → $\frac{12}{6 \times 10^{23}} \times 10^{22}=\frac{12}{60}=\frac{1}{5}$ ``\frac{12}{6 \times 10^{23}} \times 10^{22}=\frac{12}{60}=\frac{1}{5}`` = 0.2 g

#1-e-ii
Molecular formula of a compound is C₆H₁₈O₃. Find its empirical formula.

Ans : Molecular formula = C₆H₁₈O₃

Take the common multiple

Molecular formula = (C₂H₆0)₃

Molecular formula = (Empirical formula)„

Thus, empirical formula = C₂H₆O