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

Explanation

Let the weight of the mixture is w g.
So, in the mixture, the weight of `H_(2)` gas=0.2 w g and that
of `O_(2)` gas =(w-0.2w)=0.8 w g
0.2 w g of `H_(2)=(0.2w)/(2)=0.1xxw` mol of `H_(2)`
0.8 wg of `O_(2)=(0.8w)/(32)=0.025xxw` mol of `O_(2)`
Mole fraction of `H_(2)(x_(H_(2)))=(0.1w)/(0.1xxw+0.025xxw)=0.8`
So, the partial pressure of `H_(2)` in the mixture `=x_(H_(2))xx`total pressure of the mixture `=0.8xx1`bar=0.8 bar