The vapour pressure of a dilute aqueous solution of glucose $(C_{6} H_{12} O_{6})$ is 750 mm Hg. Calculate (a) molality . ( Give answer as $10^{3} \times x$ )

Open in App

Solution

Given that : temperature =273 K
Boiling point of $H_{2} O$ = 373 K
Vapour pressure of $H_{2} O$ =76 cm
We have,
$\frac{P^{o} - P_{S}}{P^{o}} = \frac{W_{2} \times M w_{1}}{M w_{2} \times W_{1}}$
$∴ M o l a l i t y = \frac{W_{2}}{M w_{2} \times W_{1}} \times 1000 = \frac{P^{o} - P_{S}}{P^{o}} \times \frac{1}{M w_{1}} \times 1000$

$= \frac{760 - 750}{760} \times \frac{1}{18} \times 1000$
=0.730 mol/kg of solvent
Also, we have
$\frac{P^{o} - P_{S}}{P^{o}} = \frac{n_{2}}{n_{2} \times n_{1}}$
$∴$ Mole fraction= $\frac{P^{o} - P_{S}}{P^{o}} = \frac{760 - 750}{760} = \frac{10}{760} = 0.013$

Suggest Corrections

Similar questions

(C_{6} H_{12} O_{6})

750 m m

373 K

750

373

750 mm Hg

373 K

(i)

(i i)

H g

C_{6} H_{12} O_{6}

100^{\circ} C

Join BYJU'S Learning Program