Question

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

Answer 1

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

Then, the number of moles of dihydrogen,  ${\displaystyle {\text{n}}_{{\text{H}}_2}=\frac{20}{2}=10}$ and the number of moles of dioxygen, ${\displaystyle {\text{n}}_{{\text{O}}_2}=\frac{80}{32}=2.5}$ moles

Given,

Total pressure of the mixture, ${\displaystyle {\text{p}}_{\text{total}}}$ = 1 bar

Then, partial pressure of dihydrogen,

${\displaystyle {\text{p}}_{{\text{H}}_2}=\frac{{\text{n}}_{{\text{H}}_2}}{{\text{n}}_{{\text{H}}_2}+\prime n\text{\_(}O\text{\_2)) xx}P\text{\_}\to tal}}$

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

= 0.8 bar

Hence, the partial pressure of dihydrogen is 0.8 bar.