Question

A solution is prepared by dissolving 394 g of nonvolatile solute in 622 g of water . The vapour pressure of solution is found to be 30 74 mm Hg at 30°C. If vapour pressure of water at 30°C is 31.8 mm Hg, what is the molar mass of solute ?

Answer 1

Explanation:

According to raoult's law,

p

0

p

0

−p

=

m×W

w×M

Where,

p

0

is Vapour pressure of pure water

p is vapour pressure of solvent

w is weight of solute

m is molecular weight of solute

W is molecular weight of solvent

32

32−31.84

=

m×200

10×18

⇒m=

0.005

0.9

=180

Answer 2

Given: A solution is prepared by dissolving 394 g of nonvolatile solute in 622 g of water. The vapour pressure of the solution is found to be 30.74 mm Hg at 30° C. The vapour pressure of water at 30° C is 31.8 mm Hg.

To find: The molar mass of the solute.

Solution:

• For a solution containing a non-volatile solute, the vapour pressure of the solution at a particular temperature is equal to the product of the vapour pressure of the solvent in the pure state and its mole fraction in the solution.

        $\frac{p^{0}-p}{p^{0}} = \frac{w * M}{m * W}$

• Here, p⁰ is the vapour pressure of pure water, p is the vapour pressure of the solvent, w is the weight of the solute, M is the molecular weight of the solvent, m is the molecular weight of the solute and W is the weight of the solvent.

        $\frac{31.8-30.74}{30.74} = \frac{394g * 18g}{m * 622 g}$

        $m = 333.80 g$

Therefore, the molar mass of the solute is 333.80 g.