Volume at \(0^{\circ} \mathrm{C}\) and 1 bar pressure \(=\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)
\(\begin{array}{ll}
\text { or, } & \frac{0.1 \mathrm{bar} \times 34.05 \mathrm{~mL}}{(546+273) \mathrm{K}}=\frac{1 \times \mathrm{V}_2}{273 \mathrm{~K}} \\
& \mathrm{~V}_2=\frac{0.1 \times 34.05 \times 273}{(819 \mathrm{~K})} \\
\text { or, } & \mathrm{V}_2=1.135 \mathrm{~mL} . \\
\therefore \quad & \text { Weight of } 1.135 \mathrm{~mL} \text { of vapour at } 0^{\circ} \mathrm{C} \text { and } 1 \mathrm{bar} \\
\text { pressure }=0.0625 \mathrm{~g} \\
\therefore \quad & \text { Weight of } 22700 \mathrm{~mL} \text { of vapour at } 0^{\circ} \mathrm{C} \text { and } 1 \mathrm{bar} \\
& \text { pressure } \\
& =\frac{0.0625}{1.135} \times 22700=1250 \mathrm{~g} \\
\therefore \quad & \text { Molar mass }=125 \mathrm{~g} \mathrm{~mol}-1
\end{array}\)