The mass of $112 c m^{3}$ of ${C H}_{4}$ gas at STP is :

0.08g

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1.6g

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0.16g

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0.8g

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Solution

The correct option is A 0.08g
$1 L = 1000 c m^{3}$

so, volume = $\frac{112}{1000} = 0.112 L$

Molar mass of $C H_{4}$ - $12 + 4 = 16$

At STP, 1 mole = $22.4 L$

It means $22.4 L$ contains $16 g$ methane.

$0.112 L$ contains = $\frac{16}{22.4} \times 0.112 = 0.08 g$

Hence, option A is the correct answer.

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