A mixture of Ar (40) and N₂ gases have a density of 1.40 g L^-1 at STP. Hence, weight fraction of N₂ approximately in the mixture is

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{n}}{\mathrm{V}}\mathrm{RT}=\frac{\mathrm{W}}{\mathrm{mV}}\mathrm{RT}=\frac{\mathrm{dRT}}{\mathrm{m}} \\ \therefore \mathrm{m}\left(\text{ molar mass of mixture }\right)=\frac{\mathrm{dRT}}{\mathrm{p}}=\frac{1.40\times 0.0821\times 273}{1} \\ =31.38\mathrm{g}{\mathrm{mol}}^{-1} \\ \text{ Let mixture contains }\mathrm{x}\text{ part }{\mathrm{N}}_2\text{ and }(1-\mathrm{x})\text{ part Ar. } \\ \therefore 31.38=\frac{{\mathrm{m}}_1\mathrm{x}+{\mathrm{m}}_2(1-\mathrm{x})}{\mathrm{x}+(1-\mathrm{x})} \\ \Rightarrow 31.38=\frac{28+40(1-\mathrm{x})}{1} \\ \therefore \mathrm{x}=0.72 \end{gathered}$$