Question

(a) $1.0\, L$ of a mixture of $CO$ and $CO_2$ 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.

Answer 1

(a) After passing through red-hot charcoal, following reaction occurs
$ C (s)+ CO_2(g ) \longrightarrow 2CO(g)$
If the $1.0 \,L$ original mixture contain $x$ litre of $CO_2$ , after passing from tube containing red-hot charcoal, the new volumes would be :
$2x$ (volume of $CO$ obtained from $CO_2) + 1$
$-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 $CO_2$ ,
i.e. $40\%\, CO$ and $60\% \,CO_2$ by volume.
(b) According to the given information, molecular formula of the compound is $M_3N2$ . 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 $