Two beaker A and B present in a closed vessel. Beaker A contains 152 .4 g aqueous solution of urea, containing 12 g of urea. Beaker B contains 196.2 g glucose solution, containing 18 g of glucose. Both solutions allowed to attain the equilibrium. Determine wt. % of glucose in it's solution at equilibrium:

Step-by-step solution

Mole fraction of urea in its solution = 12 60 12 60 + 140 . 4 18 = 0.025 Mole fraction of glucose = 18 180 18 180 + 178.2 18 = 0.01 Mole fraction of glucose is less so vapour pressure above the glucose solution will be higher than the pressure above urea solution, so some H 2 O molecules will transfer from glucose to urea side in order to make the solutions of equal mole fraction to attain equilibrium, let x moles H 2 O transferred ∴ 0.2 0.2 + 7.8 + x = 0.1 0.1 + 9.9 - x ⇒ x = 4 Now mass of glucose solution ⇒ 196.2