A vessel contains equal masses of Hydrogen and Helium. The fraction of the partial pressure of Hydrogen in the mixture is

Molar mass of Hydrogen ,MW = 2 g/mol

Molar mass of Helium, MW = 4 g/mol

Mass of ${\mathrm{H}}_2=\mathrm{Mass} \mathrm{of} \mathrm{He} =\mathrm{X}$

${\mathrm{n}}_{{\mathrm{H}}_2}=\frac{\mathrm{x}}{2}; {\mathrm{n}}_{\mathrm{He}}=\frac{\mathrm{x}}{4}$

$\mathrm{Mole}\hspace{0.17em}\mathrm{fraction} \mathrm{of} {\mathrm{H}}_2=\frac{\mathrm{No}.\mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{H}}_2}{\mathrm{Total} \mathrm{moles} \mathrm{of} \mathrm{mixture}}$Mole Fraction of ${\mathrm{H}}_2,{\mathrm{X}}_{{\mathrm{H}}_2}=\frac{{\mathrm{n}}_{{\mathrm{H}}_2}}{{\mathrm{n}}_{{\mathrm{H}}_2}+{\mathrm{n}}_{{\mathrm{H}}_{\mathrm{e}}}}$

$=\frac{\frac{\mathrm{x}}{2}}{\frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{4}}$

$\frac{\cancel{x}}{\cancel{2}}\times \frac{\cancel{4^2}}{3\cancel{\mathrm{x}}}=\frac{2}{3}$

$\mathrm{Partial}\hspace{0.17em}\mathrm{pressure} \mathrm{of} {\mathrm{H}}_2,{\mathrm{P}}_{{\mathrm{H}}_2}={\mathrm{x}}_{{\mathrm{H}}_2}\times {\mathrm{P}}_{\mathrm{total}}$

$\frac{2}{3}\times P_{total}$

Therefore, the fraction of partial pressure of hydrogen is $\frac{2}{3}$