A solution contains 72% water d 28% methyl alcohol. Calculate the mole firaction of each component in the solution.

Explanation

If solution has 72% `H_2O` by mass and 28% `CH_3OH` by mass, then
mass of `H_2O` in 100 g of solution = 72 g
and moles (n) of `H_2O=(72)/(14)=4.00` mol
mass of `CH_2OH` in 100 gm of solution = 28 g
(`therefore` Molar mass of `CH_3OH=32)`
`therefore` Moles (n) of `CH_3OH=(28)/(32)=0.88` mol
`therefore` Total moles = `4.0+0.88=4.88`
`=4.88` mol
Hence, mole fraction of `H_2O=("Moles of"H_2O)/"Total moles"`
`=(4.0)/(4.88)=0.82`
Mole fraction of water = 0.82
`therefore` Mole fraction of `CH_3OH=1-0.82=0.18`