A 10g mixture of hydrogen and helium is contained in a vessel of...

$PV = {n}_{mix}RT$

${n}_{mix} = \dfrac{6 \times 12 . 5}{0 . 083 \times 300} \approx 3$

Out of total 3 moles Let mole of $He = x$

Mole of ${H}_{2} = 3 - x$

Then total mass of the mixture will be = $4x + 2\left(3 - x\right) = 10$

Number of moles of He gas $x = 2mol$

Mass of $He = numberofmoles \times molarmass = 8g$