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

Explanation

The percent of `H_2=20`
`therefore` The percent of `O_2=80`
`n_(H_2)=W/(GMW)=0.2/2=0.1`
`n_(O_2)=0.8/32=0.025`
Mole fraction of `H_2=0.1/(0.1 times 0.025)`
`=0.1/0.125=0.8`
Partial pressure of `H_2`= Mole fraction of `H_2 times P_(t otal)=0.8 times 1=0.8 b ar`