Question

A solution is prepared by

dissolving 394 g of a nonvolatile solute

in 622 g of water. The vapour pressure of

solution is found to be 30.74 mm Hg at

30 0

C. If vapour pressure of water at 30 0

C

is 31.8 mm Hg, what is the molar mass​

Answer 1

Answer:

use relative lowering in vapour pressure concept

you can prfer ncert.

Answer 2

Answer:

Solution:

Given

Mass of nonvolatile solute (w_{2}) = 394g

Mass of water (W_{1}) = 622g

Vapour pressure of solution (P j )=30.74 mm.H

g

Vapour pressure of pure water (P_{4}) = 31.8mmHg

To find -

Molar mass of nonvolatile solute (M₂)

Formula:

Calculation: Molar mass of water is 18 g mol

Now, using the formula,

(P_{1} ^ 2 - P_{1})/(P_{2} ^ 2) = (W_{2}*M_{1})/(M_{1}*W)

(31mmHg - 30.74mmHg)/(31mHg) = (394g * 18gmo * l ^ 2)/(M_{1} * 622 * g)

0.0333 = 11.4 g mol/M.2

M_{2} = (11.4gmo * l ^ - 1)/0.033 = 342gmol-

:

The molar mass of the nonvolatile solute is 342 g mol™¹.