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

`14.49%`

Explanation

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%`