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10 " mL of " a mixture of methane, ethyl...

10 " mL of " a mixture of methane, ethylene. And carbon dioxide was exploded with excess of air. After the explosion, there was a contraction of 17 mL and after treatment with KOH, there was a further contraction of 14 mL. What was the composition of the mixture?

Text Solution

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Let he formula of the hydrocarbon be `C_xH_y`. The combustion of the hydrocarbon can be shown as:
`underset(xml)C_xH_y+underset(10(x(y)/(4))ml)(x+(y)/(4))O_2rarrunderset(10xml)xCO_2+(y)/(2)H_2O`
The first reduction in volume after explosion
`=10+10(x+(y)/(4))-10x=20`
`=10+(10y)/(4)=20`
Thus, `y=(10xx4)/(10)=4`
Volume of carbon dioxide produced `=20ml`
Thus, `10x=20`
`impliesx=(20)/(10)=2`
Hence, the molecular formula of the hydrocarbon is `C_2H_4`.
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