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0.4g of a compound on complete combustion gave 56ml of ${CO}_{2}$ at 760mm and 0°C The percentage of carbon in the compound is

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27.5

7.5

answer is D.

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Detailed Solution

P = 1atom

t = 273

V = 56 ml

PV=nRT

$1 \times 56 \times 10^{- 3} = \frac{w}{44} \times 22.4$

$w = \frac{1 \times 56 \times 10^{- 3} \times 44}{22.4}$

$w = 0.11 g$

Percentage of $C = \frac{12}{44} \times \frac{0.11}{0.4} \times 100$

=7.5%

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