Some Basic Concepts of Chemistry - Result Question 57
56. (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.
(1980,5 M)
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Answer:
Correct Answer: 56. (a) 0.6, (b) 24
Solution:
- (a) After passing through red-hot charcoal, following reaction occurs
$C(s)+CO _2(g) \longrightarrow 2 CO(g)$
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 :
$2 x$ (volume of $CO$ obtained from $\left.CO _2\right)+1$
$\begin{aligned} & \\ \Rightarrow \quad x \text { (original CO) } & =1+x=1.6 \text { (given) } \\ x & =0.6 \end{aligned}$
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 $M _3 N _2$. Also, 1.0 mole of compound has $28 g$ of nitrogen. If $X$ is the molar mass of compound, then :
$\begin{gathered} X \times \frac{28}{100}=28 \\ \Rightarrow \quad X=100=3 \times \text { Atomic weight of } M+28 \\ \Rightarrow \quad \text { Atomic weight of } M=\frac{72}{3}=24 \end{gathered}$
BETA
