Question

A solution containing 15 g of non-volatile solute dissolved in 45 g of water has a vapour pressure of 1400 Pa at 298 K. Further, 9 g of water is then added to the solution and the new vapour pressure becomes 1450 Pa. Calculate the molar mass of the solute.

Answer 1

Explanation:

hope it helps just check it

Answer 2

Given:

Mass of solute = 15 $gm$

Mass of water = 45  $gm$

Vapour pressure of solution = 1400 $Pa$

After adding 9  $gm$ water Vapour pressure of solution = 1450  $Pa$

Temperature = 298 $K$

To Find:

Molar mass of solute

Solution:

We know from Vapour pressure formula

$\dfrac{P - P_{s}}{P_{s}} = \dfrac{w}{m}\dfrac{M}{W}$

First equation

$\dfrac{P - 1400}{1400} = \dfrac{15}{m}\dfrac{18}{45} = \dfrac{6}{m}$

Second equation

$\dfrac{P - P_{s}}{P_{s}} = \dfrac{w}{m}\dfrac{M}{W + added solvent}$

$\dfrac{P - 1450}{1450} = \dfrac{15}{m}\dfrac{18}{54} = \dfrac{5}{m}$

Compare both first and second equation we will get P value

$P = 1723.07$ $Pa$

After putting P value we will get molar mass of solute

Molar mass = 26.54