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`
`because` 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.
`therefore (0.2)/(0.2 + 7.8 + x)= (0.1)/(0.1 + 9.9 -x) rArr x=4`
`therefore (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`