$11.2$ litres of a gas at STP weighs $14$ g. The gas could not be :

N_{2}

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C O

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B_{2} H_{6}

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N_{2} O

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Solution

The correct option is B $N_{2} O$
Molar mass of gas = $\frac{14 \times 22.4}{11.2} = 28 g {m o l}^{- 1}$ , and $C O$ , $N_{2}$ , $B_{2} H_{6}$ all have molar mass of $28$ while $N_{2} O$ have molar mass of $44$ .

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11.2 litre of a gas at STP weighs 14 g. The gas could not be:

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