A gaseous mixture of He and $O_2$ is found to have a density of $0.543 \mathrm{g} {\mathrm{dm}}^{-3}$at 27 °C and 760 Torr. The mass per cent of helium in the mixture is

$M_{2v}=\frac{\rho RT}{p}=\frac{\left(0.543\mathrm{g}{\mathrm{dm}}^{-3}\right)\left(8.314{\mathrm{kPadm}}^3{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\right)\left(300\mathrm{K}\right)}{\left(101.325\mathrm{kPa}\right)}=13.37\mathrm{g}{\mathrm{mol}}^{-1}$If x is the mole fraction of He, then $x{M}_{\mathrm{He}}+(1-x)M_{{\mathrm{O}}_2}=13.37\mathrm{g}{\mathrm{mol}}^{-1}$$\text{ i.e. }x\left(4\right)+(1-x)32=13.37$ 

This give  $x=0.67$

Mass of He $\begin{array}{c}=\left(4\mathrm{g}\right)0.67=2.68\mathrm{g}\end{array}$

and Mass of ${\mathrm{O}}_2=\left(32\mathrm{g}\right)0.33=10.56\mathrm{g}$

Mass per cent $\mathrm{He}=\frac{2.68}{2.68+10.56}\times 100=20.2$