30 mL of a mixture of CH4 and C2H6 was burnt in excess O2. After cooling, the CO2 formed was absorbed by KOH and the contraction on absorption was 40 mL. The mole fraction of C2H6 in the mixture was

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$30 mL$ of a mixture of $CH X_{4}$ and $C X_{2} H X_{6}$ was burnt in excess $O X_{2}$ . After cooling, the $CO X_{2}$ formed was absorbed by KOH and the contraction on absorption was $40 mL$ . The mole fraction of $C X_{2} H X_{6}$ in the mixture was

Let

CH X_{4} = a

C X_{2} H X_{6} = 30 - a

CO X_{2}

= a + 2 (30 - a) = 60 - a = 40

, so

a = 20

and

C X_{2} H X_{6} = 10 mL

. Mole fraction

= 10/30 = 1/3