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

ΔTb = Kb*M

ΔTf = Kf*M

ΔTb = Tb(solution) -Tb(pure water) = 0.303°C = 0.5 * M

⇒ M = 0.303/0.5

ΔTf = 1.87 * 0.303/0.5 = 1.133°C

ΔTf = Tf(pure water) - Tf(solution)

1.133°C = 0 - Tf(solution)

Tf(solution) = - 1.133°C

Answer 2

Δ T_b = K_bm

∵ Δ T_b = 373.303 - 373

= 0.303 k

∴ 0.303 = 0.5 m

m = \(\frac{0.303}{0.5}\)

m = 0.606

∴ Δ T_f = K_fm

= 1.87 x 0.606

Δ T_f = 1.133

⇒ T_f = (273.0 - 1.133)k

T_f = 271.867 k

or T_f = -1.133°c