Question

Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber’s salt, Na₂SO₄ . 10H₂O in 0.100 kg of water. (Given: K_f for water = 1.86 K kg mol⁻¹, atomic masses: Na = 23, S = 32, O = 16, H = 1 amu)

Answer 1

ΔT_f = i . K_f . m

Formula: Na₂SO₄ · 10H₂O

Na = 23 × 2 = 46

S = 32

O (from SO₄) = 16 × 4 = 64

10H₂O = 10 × (2 + 16) = 10 × 18 = 180

Molar Mass = 46 + 32 + 64 + 180 = 322 g/mol

Moles of solute = `6/322 = 0.01863` mol

Mass of water = 0.100 kg

Molality (m) = `0.01863/0.100 = 0.1863` mol/kg

Vant Hoff factor (i):

\[\ce{Na2SO4 -> 2Na+ + SO^2-_4}\]

i = 3

ΔT_f = i . K_f . m

= 3 × 1.86 × 0.1863

ΔT_f = 1.039 K

Then T_f  = 273.15 − 1.039

= 272.11 K

∴ The expected freezing point of the solution is 272.11 K.