Question

8.0575 × 10⁻² kg of Glauber’s salt are dissolved in water to obtain 1 dm³ of a solution of density 1077.2 kg m⁻³. Calculate the molarity, molality, and mole fraction of Na₂SO₄ in the solution.

Answer 1

Given: Mass of Glauber’s salt (Na₂SO₄.10H₂O) = 8.0575 × 10⁻² kg

= 80.575 g

Volume of solution = 1 dm³ = 1 L

Density of solution = 1077.2 kg/m³

= 1.0772 g/cm³

= 1077.2 g/L 

Molar mass of Glauber’s salt (Na₂SO₄·10H₂O) = (2 × 23) + (32) + (4 × 16) + (10 × 18)

= 46 + 32 + 64 + 180

= 322 g/mol

Moles of Na₂SO₄·10H₂O = `80.575/322`

= 0.25 mol

Since volume = 1 L

Molarity (M) = `0.25/1`

= 0.25 mol/L

Mass of solvent = Mass of solution − Mass of solute

= 1077.2 − 80.575

= 0.996625 kg

Then, Molality (m) = `0.25/0.996625`

= 0.251 mol/kg

Mole fraction of Na₂SO₄·10H₂O

Moles of water = `996.625/18`

= 55.37 mol

Total moles = 0.25 + 55.37

Total moles = 55.62

∴ Mole fraction of Na₂SO₄·10H₂O = `0.25/55.62`

= 0.0045

∴ Molarity = 0.25 mol/L, Molality = 0.251 mol/kg and Mole fraction of Na₂SO₄·10H₂O is 0.0045.