An aqueous solution of urea (mol . mass 60 g mol − 1 ) boils at 100.18 o C at 1 atm . Find the freezing temperature of soultion. Given: K f and K b for water are 1 . 86 and 0.512 K kg mol − 1 , respectively.

Step-by-step solution

Δ T f = K f m ......(1) Δ T b = K b m ......(2) ⇒ Δ T f Δ T b = K f K b ...... (3) Δ T f → Depression in freezing point Δ T b → Elevation in boiling point Boiling point of water = 100 o C Kf = 1.86 K kg mol − 1 Boiling point of urea in water = 100.18 o C Kb = 0.512 K kg mol − 1 ⇒ Δ T b = 0.18 Freezing point of water = 0 o C Freezing point of urea in water = − T o C ⇒ Δ T f = T ⇒ from equation (3), T 0.18 = 1.86 0.512 ⇒ T = 0.6539 ⇒ Freezing point urea in water = − 0 .654 o C