At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 m m H g , lowering of vapour pressure will be: (molar mass of urea = 60 g m o l - 1 )

Step-by-step solution

Relative lowering in vapour Pressure ⇒ Mole of urea n B = 0.6 60 = 10 - 2   m o l e ; Mole of water n A = 360 18 = 20 P o - P s P o = x B = P o - P s P o = n B n n + n B Here = P o = V . P . of pure solvent P s = V . P of solution. n A + n B ≃ n A P o - P s P o = n B n A l o w e r i n g   o f   v a p o u r   p r e s s u r e ( P o - P s ) = n B n A × P o P o - P s = 10 - 2 20 × 35 P o - P s = 0.0175    m m of H g