Question

Elevation in b.pt. of an aqueous glucose solution is $0.6$ $K _{ b }$ for water is $0.52\, K$ molality $^{-1} .$ The mole fraction of glucose in the solution is :

• A. 0.02
• B. 0.03
• C. 0.01
• D. 0.04

Answer 1

Answer: (A)

$\Delta T=\frac{1000 \times K_{b} \times n \times M}{W \times M}$

$=\frac{1000 \times K_{b} \times n}{N \times M}$

or $\frac{n}{N}=\frac{\Delta T \times M}{1000 \times K_{b}}$

$=\frac{0.6 \times 18}{1000 \times 0.52}=0.02$

$\therefore \frac{N}{n}=50$ or

$1+\frac{N}{n}=51$

$\therefore \frac{n+N}{n}=51 $

$ \therefore \frac{n}{n+N}=0.02$