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 for water is 0.512 K kg mol^-1 ).

Answer 1

K_b = 0.512 K kg mol^-1

K_f = 1.86 K kg mol^-1

ΔT_b = 100.18° – 100° = 0.18

ΔT = K_b x molality molality

= 0.18/0.512

Now, ΔT_f = K_f x molality

= 1.86 x \(\frac{0.18}{0.512}\) = 0.6539

Freezing point of solutions = 0 – 0.6539 = -0.6539