Let the total mass of the mixture be 100 g Mass of $\mathrm{H}_2$ = $\frac{20}{100}$ × 100 = 20 g , No. of moles of $\mathrm{H}_2$ = $\frac{20}{2}$ = 10 moles Mass of $\mathrm{O}_2$ = 100 – 20 = 80 g, No. of moles of $\mathrm{O}_2$ = $\frac{80}{32}$ = 2.5 moles Total number of moles of $\mathrm{H}_2$ and $\mathrm{O}_2$ = 10 + 2.5 = 12.5 Mole fraction of $\mathrm{H}_2$ = $\frac{10}{12.5}$ = $\frac{4}{5}$ = 0.8 Total pressure of mixture = 1 bar ∴ Partial pressure of $\mathrm{H}_2$ in the mixture, $p_{\mathrm{H}_2}$ = Mole fraction × Total pressure = 0.8 × 1 = 0.8 bar