Question

Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in the weight of solution was 2.50 g and that of pure solvent was 0.04 g. The molecular weight of the solute is

Answer 1

Let P° = partial pressure of pure solvent

Ps = Partial pressure of solution.

Ps / P° = 2.5 / 2.54

(P° - Ps) / Ps = 0.04 / 2.5 = 0.016

(P° - Ps) / Ps = moles of solute / moles of solvent

Let. Molar mass of solute = m

Moles of water equals :

180 / 18 = 10moles

0.016 = (5 / m) / (10)

0.016 = 5 / 10m

0.16 m = 5

m = 5 / 0.16

m = 31.25 grams / mol

Answer 2

Let P° = partial pressure of pure solvent

Ps = Partial pressure of solution.

Ps / P° = 2.5 / 2.54

(P° - Ps) / Ps = 0.04 / 2.5 = 0.016

(P° - Ps) / Ps = moles of solute / moles of solvent

Let. Molar mass of solute = m

Moles of water equals :

180 / 18 = 10moles

0.016 = (5 / m) / (10)

0.016 = 5 / 10m

0.16 m = 5

m = 5 / 0.16

m = 31.25 grams / mo