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Ten millilitre of a gaseous hydrocarbon ...

Ten millilitre of a gaseous hydrocarbon is was exploded with oxygen. After the explosion, there was a contraction of 20 ml in volume. On shaking the residual gaseous mixture with `KOH`, there was a further concentration of 20 ml in volume. Calculate the molecular formula. All the volumes were recorded at same temperature and pressure.

Text Solution

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Let the formula of the hydrocarbon be `C_(x)H_(y)`.
The combustion of the hydrocarbon can be shown as:
`C_(x)H_(y) + (x+(y)/(4))O_(2) rarr xCO_(2)+(y)/(2)H_(2)O`
`10mL 10(x+(y)/(4))mL 10x mL`
The first reduction in volume after explosion
`= 10 +10 (x+(y)/(4)) - 10 x = 20`
`=10 +(10y)/(4) = 20`
Thus, `y = (10 xx 4)/(10) = 4`
Volume of carbon dioxide produced `= 20mL`
Thus, `10 x = 20`
`x = (20)/(10) = 2`
Hence, the molecular formula of the hydrocarbon `= C_(2)H_(4)`
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