Question

Determine the freezing point of a solution containing 0.625 g of glucose (C₆H₁₂O₆) dissolved in 102.8 g of water. 

(Freezing point of water = 273 K, K_f for water = 1.87 K kg mol⁻¹, at. wt. C = 12, H = 1, O = 16)

Answer 1

Given: Mass of glucose = 0.625 g

Molar mass of glucose (C₆H₁₂O₆) = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g/mol

Mass of water = 102.8 g = 0.1028 kg

K_f = 1.87 K kg mol⁻¹

Freezing point of pure water = 273 K

Moles of glucose = `0.625/180`

≈ 0.00347

Molality (m) = `(0.00347 mol)/(0.1028 kg)`

≈ 0.03375 mol/kg

We know that

ΔT_f ​= K_f​ m

= 1.87 × 0.03375

≈ 0.063 K

T_f ​= 273 − ΔT_f ​

= 273 − 0.063

= 272.94 K

∴ The freezing point of the solution is 272.94 K.