Question

The density of a gas is $1.27g{L}^{-1}$ at ${50}^{\circ }C$ and $0.987\times {10}^{5}Pa$. Calculate its molar mass.

Answer 1

We know that the equation for an ideal gas is:

$\begin{array}{c}PV=nRT\\ PV=\frac{m}{M}RT\\ M=\frac{mRT}{VP}=\frac{\rho RT}{P}\end{array}$

where $\rho$ is the density of gas.

Given

$\begin{array}{c}\rho =1.27g{L}^{-1}\\ T={50}^{\circ }C=50+273=323K\\ P=0.987\times {10}^{5}Pa=0.987atm\\ R=0.0082057Latmmo{l}^{-1}{K}^{-1}\\ M=\frac{1.27\times 0.082057\times 323}{0.987}=34.103\simeq 34g\end{array}$