2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure

Volume (V) occupied by dihydrogen is given by,

$V=\frac{m}{M} \frac{\mathrm{R} T}{p}$

$=\frac{0.184}{2} \times \frac{\mathrm{R} \times 290}{p}$

Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:

$V=\frac{m}{M} \frac{\mathrm{R} T}{p}$

$=\frac{2.9}{M} \times \frac{\mathrm{R} \times 368}{p}$

According to the question,

$\frac{0.184}{2} \times \frac{\mathrm{R} \times 290}{p}=\frac{2.9}{M} \times \frac{\mathrm{R} \times 368}{p}$

$\Rightarrow \frac{0.184 \times 290}{2}=\frac{2.9 \times 368}{M}$

$\Rightarrow M=\frac{2.9 \times 368 \times 2}{0.184 \times 290}$

$=40 \mathrm{~g} \mathrm{~mol}^{-1}$

Hence, the molar mass of the gas is $40 \mathrm{~g} \mathrm{~mol}^{-1}$.