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

Explanation

Suppose, Weight of mixture gas = 100 g
`therefore` Weight of `H_(2)=20 g` and
Weight of `O_(2)=80 g`
`therefore` Mole of `n_(H_(2))=(20)/(2)=10` mol and
Mole of `n_(O_(2))=(80)/(32)=2.5` mol
Total mole of mixture gas `= (10+2.5)=12.5` mol
Mole fraction of `H_(2), chi_(H_(2))=(n_(H_(2)))/("Total Moles")=(10)/(12.5)`
Mole fraction of `O_(2), chi_(O_(2))=(n_(O_(2)))/("Total moles")=(2.5)/(12.5)`
Partial Pressure = Mole fraction `xx` Total pressure
`therefore p_(H_(2))=n_(H_(2))xx P=(10xx"1 bar")/(12.5)=(10xx"1 bar")/(12.5)=0.8` bar
`therefore p_(O_(2))=n_(O_(2))xxP=(2.5xx"1 bar")/(12.5)=0.2` bar