2.17   The vapour pressure of water is 12.3 \; k Pa at 300 \; K.  Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answers (1)

It is asked the vapour pressure of 1 molal solution which means 1 mol of solute in 1000 g H2O.

Moles in 1000g of water = 55.55 mol.         (Since the molecular weight of H2O is 18)

Mole fraction of solute :

                                                 \frac{1}{1+55.55} = 0.0177

Applying the equation :

                                          \frac{p_w^{\circ} - p}{p_w^{\circ}} = x_2

or                                       \frac{12.3 - p}{12.3} = 0.0177

or                                            p = 12.083\ KPa

Thus the vapour pressure of the solution is 12.083 KPa

Related Chapters

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions