Two beakers A and B are present in a closed vessel. Beaker A contains 152.4 g aqueous solution of urea containing 12 gofurea. Beaker B contains 196.2 g glucose solution, containing 18 gof glucose. Both solutions are allowed to attain equilibrium. Determine the w.t.% of glucose in it's solution at equilibrium:

Options

A. 6.71
B. 14.49
C. 16.94
D. 20

Answer

B

Explanation

Mole fraction of urea in it's solution `=((12)/(60))/((12)/(60)+(140.4)/(18)) rArr 0.025`
Mole fraction of glucose `=((18)/(180))/((18)/(180)+(178.2)/(18)) rArr 0.01`
Mole fraction of glucose is less than that of urea. Therefore, vapour pressure above 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) rArr x=4`
Mass of glucose solution `rArr 196.2 -4 xx 18 rArr 124.2` Wt. of glucose `=(18)/(124.2) xx 100=14.49`