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Nine millilitre of a mixture of methane ...

Nine millilitre of a mixture of methane and ethylene was exploded with 30 ml (excess) of oxygen. After cooling, the volume was 21.0 ml. Further treatment with caustic potash solution reduced the volume to 7.0 ml. Determine the volume of `CH_4` in the mixture.

Text Solution

Verified by Experts

The correct Answer is:
4
Let the valume of `CH_4` = xml
Volume of `(CH_2=CH_2) = (9-x)ml`
(i) `underset(xml)(CH_4)(g) + underset(2xml)(2O_2(g))rarrunderset("xml")(CO_2(g))+2H_2O(l)`
ii) `underset((9-x)xml)(CH_2)=CH_2(g) +underset("3(9-x)ml)(3O_2(g))rarrunderset(2(9-x))(2CO_2)(g)=H_2O(l)`
Volume of `CO_2` absorbed by KOH = 21 - 7 = 14 ml
`:. x +2 (9 - x) = 14 ml`
`:. x = 4 ml`
Volume of `CH_4=4ml`
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