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

Explanation

Let the mass of `H_(2)` in the mixture `= 20 g`
`:.` The mass of the `O_(2)` in the mixture will be `= 80 g`
No . Of moles of `H_(2) (n_(H_(2))) = ((20 g))/((2 g mol^(-))) = 10 mol`
No. of moles of `O_(2) (n_(O_(2))) = ((80 g))/((32 mol^(-1))) = 2.5 mol`
Total pressure of the gaseous mixture `= 1"bar"`
All other conditions being the same, the partial pressures of the gases are inthe ratio of their moles
Thus, partial pressure of `H_(2) (p_(H_(2))) = ((80 g))/((32 g mol^(-1))) xx (1 "bar") = 0.8 "bar"`