Example 2.17: How many liters of pure methanol and of pure water must be mixed to form two liters of antifreeze at 25 °C? Partial molar volumes of methanol and water in a 30 mol % methanol solution are:

Methanol (1): V₁' = 38.633 cm³/mol, V₁ = 40.277 cm³/mol, x₁ = 0.3

Water (2): V₂' = 17.765 cm³/mol, V₂ = 18.068 cm³/mol, x₂ = 0.7

Solution V^T = 2000 cm³

$ Specific volume of solution, V = x₁V₁' + x₂V₂' = 24.025 cm³/mol

$ Number of moles of solution, n = V^T/V = 83.246 mol

$ Number of moles of methanol, n₁ = x₁n = 24.974 mol

$ Number of moles of water, n₂ = x₂n = 58.272 mol

$ Total volume of methanol, V₁^T = n₁V₁ = 1017 cm³

$ Total volume of water, V₂^T = n₂V₂ = 1053 cm³