Question

Elevation in boiling point of an aqueous glucose solution is 0.6 $K_b\left(water\right)=0.52K{\text{molality}}^{-1})$. The mole fraction of glucose in the solution is:

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

Answer 1

The correct option is A 0.02

$\Delta T=\frac{1000\times K_b\times n\times M}{W\times M}\Delta T=\frac{1000\times K_b\times n}{N\times M}$

$or\frac{n}{N}=\frac{\Delta T\times M}{1000\times K_b}\frac{n}{N}=\frac{0.6\times 18}{1000\times 0.52}=0.02$

$\therefore \frac{N}{n}=50or1+\frac{N}{n}=51$

$\Rightarrow 1+\frac{N}{n}=51$
$\Rightarrow \frac{n}{n+N}=0.02$
Thus the mole fraction of glucose in the solution is 0.02.