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

Explanation

Given 20% by wt of dihydrogen so 80% oxygen remained for dihydrogen
` n = ( w t)/( Gm w t) = ( 0.2 )/(2) = 0.1`
for dioxygen
` n= ( w t)/( GM w t)= ( 0.8 )/( 32 ) = 0.025`
mole fraction of ` H_2 `
` = ( 0.1 )/( 0.1 + 0.025 ) = ( 0.1 )/( 0.125 ) =0.8`
Partial pressure of dihydrogen
= molefraction of `H_2 xx P_(" total ")`
` = 0.8 xx 1 = 0.8 ` bar