Question

Given K_b and K_f of water are 0.52 and 1.86 K kg mol^-1 respectively. An aqueous solution freezes at - 0.186°C, what is the boiling point of solution?

Options

• 100.052° C

• 100.52° C

• 0.52° C

• 0.052° C

Answer 1

100.052° C

Explanation:

K_b = 0.52 K kg mol^-1 

Δ T_f = K_f × m

Molality = `0.186/1.86`

Δ T_f = 0.186° C

m = 0.1

Δ T_b = K_b × m

= 0.52 × 0.1 = 0.052

Δ T_b = T'_b - T_b = 0.052 = T'_b - 100 = 0.052

Boiling point of solution T'_b = 100.052° C