Question

A solution of urea in water has a boiling point of 100.18°C. Calculate the freezing point of the solution. (K_f for water is 1.86 K kg mol⁻¹ and K_b for water is 0.512 K kg mol⁻¹).

Answer 1

Given: Boiling point of solution = 100.18°C

Boiling point of pure water = 100.00°C

ΔT_b = 100.18 − 100.00 = 0.18°C

K_b = 0.512 K kg mol⁻¹

K_f = 1.86 K kg mol⁻¹

ΔT_b = K_b m

m = `(Delta T_b)/K_b`

= `0.18/0.512`

= 0.3516 mol/kg

ΔT_f = K_f  m

= 1.86 × 0.3516

= 0.654°C

T_f = 0°C − ΔT_f

= 0 − 0.654

=  − 0.654°C

∴ Freezing point of the solution is − 0.654°C.