Question

A certain aqueous solution boils at 100.303°C. What is its freezing point? K_b for water = 0.5 K kg mol⁻¹ and K_f = 1.87 K kg mol⁻¹?

Answer 1

Given: Boiling point of solution = 100.303°C

Boiling point of pure water = 100.0°C

ΔT_b = 100.303 − 100.0 = 0.303°C

K_b = 0.5 K kg mol⁻¹

K_f = 1.87 K kg mol⁻¹

ΔT_b ​= K_b​ m

m = `(Delta T_b)/K_b`

= `0.303/0.5`

= 0.606 mol/kg

ΔT_f​ = K_f​ m

= 1.87 × 0.606

≈ 1.133°C

T_f ​= 0.0°C − ΔT_f​

= 0 − 1.133

T_f = −1.133

Freezing point of the solution is −1.133.