Question

Two beakers a and b are present in a closed vessel. Beaker a contains 152.4 gm aqueous answer of urea, containing 12 g of urea. Beaker b contain

Answer 1

Answer:

Weight %age of C6H12O6 = 18 / 124.2 x 100 = 14.49

Explanation:

Mole fraction of CO(NH2)2 = 12/60 ÷ 12/60 + ( 140.4/18) = 0.025

Mole fraction of C6H12O6 = 18/180 ÷ 18/180 + 178.2/18 = 0.01

As in case of glucose, mole fraction is lesser then the pressure above glucose will be higher than urea solution. Let x moles of water transferred from glucose to urea to make them equal.

0.2 ÷ 0.2 + 78 + x = 0.1 / 0.1 + 9.9 - x

x = 4

Mass of C6H12O6 = 196.2 - 4 x 18 = 124.2

Weight %age of C6H12O6 = 18 / 124.2 x 100 = 14.49