Question

Elevation in b.pt. of an aqueous glucose solution is $0.6$ $K_b$ for water is $0.52K$ 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

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

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

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