One litre of a certain gas weight $1.16 g$ at STP. The gas may possibly be:

C_{2} H_{2}

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C O

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O_{2}

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N H_{3}

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Solution

The correct option is A $C_{2} H_{2}$
At STP,

$\frac{V o l u m e o f a g a s}{22.4} = \frac{W e i g h t}{M o l e c u l a r w e i g h t}$

$\frac{1}{22.4} = \frac{1.16}{M o l e c u l a r w e i g h t}$

Molecular weight = $1.16 \times 22.4$

= $26 g r a m / m o l e$

Therefore, the answer is $C_{2} H_{2}$ .

Hence, option $A$ is correct.

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