Question

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.

Answer 1

Number of moles of H₂O (n₁) = {1000 g}/{18 g mol^-1} = 55.55 mol.

Mole fraction of C₂H₅OH xC₂H₅OH = 0.04

xC₂H₅OH = {n₂}/{n₁ + n₂} = {n₂}/{55.55 + n₂} = 0.04

n₂ - 0.04 n₂ = 55.55 x 0.04 = 2.222

0.96 n₂ = 2.222

n₂ = {2.222}/{0.96} = 2.31 mol

∴ Molarity of C₂H₅OH solution = 2.31 mol

Molarity of solution (M) = {No. of gram moles of C₂H₅OH}/{Volume of solution in litres}

In solution mass of H₂O = 1000 g

Mass of C₂H₅OH = (2.31 mol) x (46 g mol^-1) = 106.26 g

Total mass of the solution = 1000 + 106.26 = 1106.26 g

Density of ethanol = 0.800 g cm^-3

Volume of the solution = {1106.26 g}/{0800 g cm^-3} = 1382.8 cm³

Molarity of solution = {2.31 mol}/{(1382.8/1000) dm³} = {2.31 x 1000}/{1382.8} = 1.67 M