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A 12 gm sample of ${CH}_{4}$ and $C_{2} H_{4}$ on complete oxidation with $O_{2}$ forms 35.2 gm. Find the mean molar mass of original sample :

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23.0

14.7

answer is A.

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Detailed Solution

$\underset{a}{{CH}_{4}} + 2 O_{2} \to {CO}_{2} + 2 H_{2} O$

$\underset{b}{C_{2} H_{4}} + 3 O_{2} \to 2 {CO}_{2} + 2 H_{2} O$

$\begin{array}{l} a + b = 0.8 \\ 16 a + 28 b = 12 \end{array}$

$\begin{aligned} \Rightarrow a = 0.4 \\ b = 0.2 \end{aligned}$

$\frac{0.4 \times 16 + 0.2 \times 28}{0.6} = M_{a v g}$

Thus mean molar mass = 20.

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