Question 15

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer

Here,

Vapour pressure of the solution at normal boiling point (p₁) = 1.004 bar (Given)

Vapour pressure of pure water at normal boiling point (p₁⁰) = 1.013 bar

Mass of solute, (w₂) = 2 g

Mass of solvent (water), (w₁) = 100 - 2 = 98 g

Molar mass of solvent (water), (M₁) = 18 g mol^{- 1}

According to Raoult's law,

(p₁⁰ - p₁) / p₁⁰= (w₂x M₁) / (M₂x w₁ )

(1.013 - 1.004) / 1.013 = (2 x 18) / (M₂ x 98 )

0.009 / 1.013 = (2 x 18) / (M₂ x 98 )

M₂ = (2 x 18 x 1.013) / (0.009 x 98)

M₂ = 41.35 g mol^{- 1}

Hence, the molar mass of the solute is 41.35 g mol^{- 1}.

10 Comment(s) on this Question

Reena Roy 2019-07-18 16:08:54

Very helpful

Anupama 2019-07-06 17:49:54

In the question,it's only given as a solvent.Why did u take it as water?

Veer Pratap Singh Laungwal 2019-05-30 10:16:09

Thank you so much very informative

Satya prakash satapathy 2019-05-26 08:25:27

Thanks

Pooja 2019-05-22 01:01:14

Thx

Arushi 2019-05-05 20:41:39

Thanks

pratap 2019-05-05 08:52:22

isme vapour presure water ka to diya hi na h

Kr shiddartha 2019-04-23 22:24:47

Thanks

Prashant 2019-04-21 11:36:46

Thanks

Mayank 2019-04-02 20:40:40

Thank u so much for this help and God bless you