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The molar mass of $N_{2} O$ as well as ${CO}_{2}$ is 44 g ${mol}^{- 1}$ . At $25^{o} C$ and 1 atm pressure, 1 L $N_{2} O$ contains n molecules of gas. The number of ${CO}_{2}$ molecules in 2 L under same conditions will be:

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$\frac{n}{4}$

$\frac{n}{2}$

answer is B.

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Detailed Solution

$\frac{V_{1}}{V_{2}} = \frac{n_{1}}{n_{2}}$

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