The vapour pressure of water is 12.3 kPa | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Question:

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer:

1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).

Molar mass of water = 18 g mol - 1

∴ Number of moles present in 1000 g of water = 1000/18

= 55.56 mol

Therefore, mole fraction of the solute in the solution is

x2 =  1 / (1+55.56) = 0.0177.

It is given that,

Vapour pressure of water, p10 = 12.3 kPa

Applying the relation, (P10 P1) / P10 = X2 

⇒ (12.3 - p1) / 12.3 =0.0177

⇒ 12.3 - P1 = 0.2177

p1 = 12.0823

= 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.


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Comments

  • Ridham Rana
  • 2019-05-28 10:44:24

Thank you for help SaralStudy.


  • Bhadawar
  • 2019-04-04 20:46:15

Thanx for help


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 17: The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of ....