$10$ mL of gaseous hydrocarbon on combustion gives $40$ mL of $C O_{2} g$ and $50$ mL $H_{2} O$ (vapour). The hydrocarbon is:

C_{4} H_{5}

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C_{8} H_{10}

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C_{4} H_{8}

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C_{4} H_{10}

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Solution

The correct option is D $C_{4} H_{10}$
The ratio of volumes is $= 10 : 40 : 50$ $= 1 : 4 : 5$

All the carbon from the hydrocarbon is in the ${C O}_{2}$ . $10 m l$ of hydrocarbon produces $40 m l$ of ${C O}_{2}$ $1 : 4$ ratio.

So, $1$ mole of hydrocarbon has four moles of carbon.

All the $H$ ends up in the water.

$10 m l$ of hydrocarbon produces $50 m l$ water $1 : 5$ ratio,

Each water molecule has $2$ moles of $H$ atoms. So, $1$ mole hydrogen carbon compound has $10$ moles of hydrogen. Which gives $C_{4} H_{10}$ .

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