A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen.

Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.

Then, the number of moles of dihydrogen, $n_{\mathrm{H}_{2}}=\frac{20}{2}=10$ moles and the number of moles of dioxygen, $n_{\mathrm{O}_{2}}=\frac{80}{32}=2.5$ moles.

Given,

Total pressure of the mixture, ptotal = 1 bar

Then, partial pressure of dihydrogen,

$p_{\mathrm{H}_{2}}=\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{H}_{2}}+n_{\mathrm{O}_{2}}} \times P_{\text {total }}$

$=\frac{10}{10+2.5} \times 1$

$=0.8 \mathrm{bar}$

Hence, the partial pressure of dihydrogen is.