10.0 ml mixture of methane and ethylene was exploded with 30ml (excess) of oxygen. After cooling the volume was 21.0 ml. On treatement with KOH solution the volume was reduced to 7.0 ml. The amounts of methane and ethylene in the mixture respectively are

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5 ml methane +5 ml $C_{2} H_{4}$

3 ml $C H_{4}$ + 7 ml $C_{2} H_{4}$

6 ml $C H_{4}$ + 4ml $C_{2} H_{2}$

4 ml $C H_{4}$ + 6ml $C_{2} H_{4}$

answer is C.

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Detailed Solution

Let the vol. of $C H_{4}$ be = a ml

Vol. of $C_{2} H_{4} = (10 - a) m l$

Equation for combustion of $C H_{4}$ and $C_{2} H_{2}$

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O (l i q u i d)$

a ml 2a ml a ml negligible volume

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$

(10 – a) ml 3(10 – a)ml 2(10 – a) ml negligible

Vol. of $C O_{2}$ produced $+ O_{2}$ left = 21 ml

Vol. of $C O_{2}$ produced $= 21 - 7 = 14 m l$

$C O_{2}$ produced is

$a + 2 (10 - a) = 14$

$∴ a = 6$

$∴ V o l . o f C H_{4} = 6 m l a n d v o l . o f C_{2} H_{4} = 4 m l$

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