Two beaker A and B present in a closed vessel. Beaker A contains 152.4gaqueous solution of urea, containing12g of urea. Beaker B contains 196.2g glucose solution, containing 18g of glucose. Both solution allowed to attain the equilibrium. Determine wt. % of glucose in it's solution at equilibrium:

Two beaker A and B present in a closed vessel. Beaker A contains 152.4

1.	Two beaker A and B present in a closed vessel. Beaker A contains 152.4gaqueous solution of urea, containing12g of urea. Beaker B contains 196.2g glucose solution, containing 18g of glucose. Both solution allowed to attain the equilibrium. Determine wt. % of glucose in it's solution at equilibrium:
Answer» Solution :Beaker A : Mole fraction of urea `= ((12)/(60))/((12)/(60)+(140.4)/(18)) = (0.2)/(0.2+7.8) = 0.025` Beaker B : Mole fraction of GLUCOSE `= ((18)/(180))/((18)/(180)+(178.2)/(18)) implies 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 mole of `H_(2)O` transfered `(0.2)/(0.2+7.8+x)=(0.1)/(0.1+9.9-x) implies x=4` now mass of glucose solution `= 196.2-18xx4=124.2` `WT. %` of glucose `= (18)/(124.2)xx100` `implies 14.49%`