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

Question

The vapour pressure of a dilute aqueous solution of glucose is 750 mm of Hg at 373 K. The mole fraction of solute is: (A) (1/10) (B) (1/7.6) (C) (1/35) (D) (1/76)

Tardigrade Admin · Accepted Answer

( P 0- P s / P 0)=( n / n + N ) (At 373 K V.P. of H 2 O =760 mm Hg ) (n/n+N)=(760-750/760)=(1/76)

Q. The vapour pressure of a dilute aqueous solution of glucose is $750 mm$ of $H g$ at $373 K$ . The mole fraction of solute is:

$\frac{1}{10}$

$\frac{1}{7.6}$

$\frac{1}{35}$

$\frac{1}{76}$

$\frac{P ^{0} - P _{s}}{P ^{0}} = \frac{n}{n + N}$
(At $373 K$ V.P. of $H_{2} O = 760 mm H g$ )
$\frac{n}{n + N} = \frac{760 - 750}{760} = \frac{1}{76}$

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

101​

7.61​

351​

761​