Question

Calculate the mole fraction of water in a mixture of 12 g of water, 108 g of acetic acid and 92 g of ethyl alcohol.

Answer 1

Given: Mass of water = 12 g

Mass of acetic acid = 108 g

Mass of ethyl alcohol = 92 g

Molar mass of water (H₂O) = 18 g/mol

Molar mass of acetic acid (CH₃COOH) = 60 g/mol

Molar mass of ethyl alcohol (C₂H₅OH) = 46 g/mol

Moles of water = `12/18` = 0.667 mol

Moles of acetic acid = `108/60` = 1.8 mol

Moles of ethyl alcohol = `92/46` = 2 mol

Total moles = 0.667 + 1.8 + 2 = 4.467 mol

`X_"water" = "Moles of water"/"Total moles"`

= `0.667/4.467`

= 0.15 

∴ The mole fraction of water in a mixture is 0.15.