Ncert Exercises & Solutions

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

Let the weight of dihydrogen be 20g and the weight of dioxygen be 80g.
Then, the number of moles of dihydrogen,

n_{H_{2}} = \frac{20}{2} = 10 m o l e s

and the number of moles of dioxygen,

n_{O_{2}} = \frac{80}{32} = 2.5 m o l e s

Given, Total pressure of the mixture,

p_{t o t a l} = 1 b a r

Then, partial pressure of dihydrogen,

p_{H_{2}} = \frac{n_{H_{2}}}{n_{H_{2}} + n_{O_{2}}} \times P_{t o t a l}

= \frac{10}{10 + 2.5}

= 0.8 b a r

Hence, the partial pressure of dihydrogen is 0.8 bar.