$20 mL$ of a mixture of $CO$ and $CH X_{4}$ was exploded with $30 mL$ of $O X_{2}$ . After cooling, the residual gas measured $28 mL$ , which on treatment with KOH solution shrank to $8 mL$ . The volume of $CH X_{4}$ in the original mixture was

Answer: B

Let

CO = a

CH X_{4} = b

a + b = 20

2 CO + O X_{2} 2 CO X_{2}

and

CH X_{4} + 2 O X_{2} CO X_{2} + 2 H X_{2} O (l)

CO X_{2}

formed

= a + b = 20 mL

, absorbed by KOH:

28 - 8 = 20

, consistent. Residual

O X_{2}

= 8 = 30 - (a /2 + 2 b)

, so

a /2 + 2 b = 22

. With

a = 20 - b

10 - b /2 + 2 b = 22

b = 8 mL