Question

The vapour pressure of dilute aqueous solution of glucose is 750 mm of Hg at 373 K. The mole fraction of glucose is :

• A. $\frac{1}{10}$
  
• B. $\frac{1}{7.6}$
  
• C. $\frac{1}{35}$
• D. $\frac{1}{76}$

Answer 1

Answer: (D)

Given that:

373 K is the boiling point of $H_{2}O$ so $P^{\circ }=760$ mm of Hg

$P^o=760mm$
$P_s=750mm$

As we know,

$\frac{P^o-P_s}{P^o}=Molefraction$

Therefore, $\frac{P^{\circ }-P^s}{P^{\circ }}=X_{solute}$

$X_{solute}=\frac{760-750}{760}=\frac{1}{76}$

Therefore, the mole fraction of glucose is $\frac{1}{76}$.

Hence, option (B) is correct.