Question

2.9 g of a gas at ${95}^{\circ }C$ occupied the same volume as 0.184 g of dihydrogen at ${17}^{\circ }C$, at the same pressure. What is the molar mass of the gas?

Answer 1

Volume (V) occupied by dihydrogen is given by,
$V=\frac{m}{M}\frac{RT}{p}=\frac{0.184}{2}\times \frac{R\times 290}{p}$
Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:
$V=\frac{m}{M}\frac{RT}{p}=\frac{2.9}{M}\times \frac{R\times 368}{p}$
According to the question,
$\frac{0.184}{2}\times \frac{R\times 290}{P}=\frac{2.9}{M}\times \frac{R\times 368}{P}\Rightarrow \frac{0.184\times 290}{2}=\frac{2.9\times 368}{M}\Rightarrow M=\frac{2.9\times 368\times 2}{0.184\times 290}=40gmo{l}^{-1}$
Hence, the molar mass of the gas is $40gmo{l}^{-1}$.