If $3.7 g$ of a gas at $25^{\circ} C$ occupies the same volume as $0.814 g$ of hydrogen at $17^{\circ} C$ and at the same pressure, then what is the molecular weight of the gas?

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$P V = n R T$
$P$ and $V$ are the same.
So, $n_{1} T_{1} = n T_{2}$
or $\frac{w_{1}}{M_{1}} T_{1} = \frac{w_{2}}{M_{2}} T_{2}$
or $\frac{3.7}{M_{1}} \times 298 = \frac{0.184}{2} \times 290$
or $M_{1} = 41.33 g$

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Updated on:21/07/2023

Class 11 CHEMISTRY STATES OF MATTER

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