Question

The density of 3M aqueous solution of sodium thiosulphate is 1.25 g/ml. Calculate

1. mole fraction of sodium thiosulphate
2. molalities of Na⁺ and \[\ce{S2O^2-_3}\] ions.

Answer 1

Given: Molarity (M) of sodium thiosulphate Na₂S₂O₃ = 3 mol/L

Density of solution = 1.25 g/mL = 1250 g/L

Molar mass of Na₂S₂O₃ = 158 g/mol

i. Mass of solution for 1 L:

Volume = 1 L = 1000 mL

Density = 1.25 g/mL

Mass of solution = 1.25 × 1000 = 1250 g

Mass of Na₂S₂O₃ (solute) = 3 mol × 158 g/mol = 474 g

Mass of water (Solvent) = 1250 − 474 = 776 g = 0.776 kg

Mole fraction of Na₂S₂O₃ = `3/(3 + 43.11)`

= `3/46.11`

= 0.0651

ii. Each mole of Na₂S₂O₃ dissociates into 2Na⁺ and 1 \[\ce{S2O^2-_3}\] ion

\[\ce{Na2S2O3_{(aq)} -> 2Na+ + S2O^2-_3}\]

In 3 mol of Na₂S₂O₃, you get:

Na⁺ = 3 × 2 = 6 mol

\[\ce{S2O^2-_3}\] = 3 mol

Solvent mass = 0.776 kg

Molality of Na⁺ = `6/0.776`  = 7.73 mol/kg 

Molality of \[\ce{S2O^2-_3}\] = `3/0.776` = 3.87 mol/kg