(a) 1.0 L of a mixture of CO and CO2 is taken. This mixture is passed through a tube containing red hot charcoal. The volume now becomes 1.6 L. The volumes are measured under the same conditions. Find the composition of mixture by volume. (b) A compound contains 28 per cent of nitrogen and 72 per cent of a metal by weight. 3 atoms of metal combine with 2 atoms of nitrogen. Find the atomic weight of metal.

Question

(a) 1.0 L of a mixture of CO and CO2 is taken. This mixture is passed through a tube containing red hot charcoal. The volume now becomes 1.6 L. The volumes are measured under the same conditions. Find the composition of mixture by volume. (b) A compound contains 28 per cent of nitrogen and 72 per cent of a metal by weight. 3 atoms of metal combine with 2 atoms of nitrogen. Find the atomic weight of metal. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(a) After passing through red-hot charcoal, following reaction occurs

C (s) + C O_{2} (g) ⟶ 2 CO (g)

If the

1.0 L

original mixture contain

x

litre of

C O_{2}

, after passing from tube containing red-hot charcoal, the new volumes would be :

2 x

(volume of

CO

obtained from

C O_{2}) + 1

- x

(original

CO) = 1 + x = 1.6

(given)

\Rightarrow x = 0.6

Hence, original

1.0 L

mixture has

0.4 L CO

and

0.6 L

of

C O_{2}

,
i.e.

40% CO

and

60% C O_{2}

by volume.
(b) According to the given information, molecular formula of the compound is

M_{3} N 2

. Also,

1.0

mole of compound has

28 g

of nitrogen. If

X

is the molar mass of compound, then :

X \times \frac{28}{100} = 28

\Rightarrow X = 100 = 3 \times

Atomic weight of

M + 28

\Rightarrow

Atomic weight of

M = \frac{72}{3} = 24