(B) Let $A$ be benzene and $B$ be toluene. Given: $P_A^o = 119 \ torr$,$P_B^o = 37.0 \ torr$,$x_B = 0.50$. Since $x_A + x_B = 1$,$x_A = 0.50$. Accordi

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) Let $A$ be benzene and $B$ be toluene. Given: $P_A^o = 119 \ torr$,$P_B^o = 37.0 \ torr$,$x_B = 0.50$. Since $x_A + x_B = 1$,$x_A = 0.50$. According to Raoult's Law,the partial pressures are: $P_A = P_A^o x_A = 119 \times 0.50 = 59.5 \ torr$ and $P_B = P_B^o x_B = 37.0 \times 0.50 = 18.5 \ torr$. The total pressure $P_{total} = P_A + P_B = 59.5 + 18.5 = 78.0 \ torr$. In the vapour phase,the mole fraction of toluene $(y_B)$ is given by Dalton's Law: $y_B = \frac{P_B}{P_{total}} = \frac{18.5}{78.0} \approx 0.237$.

Class 12 Chemistry Solutions Lowering of vapour pressure

Difficult

Vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature. The mole fraction of toluene in the vapour phase,which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$,will be:

JEE MAIN 2013 All JEE MAIN

A
$0.137$
B
$0.237$
C
$0.435$
D
$0.205$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solutions

Subtopic

Lowering of vapour pressure

Previous Year

JEE MAIN 2013 Paper All JEE MAIN years →

The relative lowering in vapour pressure of an aqueous solution of urea which is $6 \%$ by weight,is

Medium

View Solution

The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm \ Hg$ and $700 \ mm \ Hg$ respectively at $350 \ K$. Find out the composition of the liquid mixture if total vapour pressure is $600 \ mm \ Hg$. Also,find the composition of the vapour phase.

View Solution

An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water,the rate at which water molecules :

MediumJEE MAIN 2020

View Solution

The vapour pressure of two liquids $P$ and $Q$ are $80$ and $60 \, torr$,respectively. The total vapour pressure of the solution obtained by mixing $3 \, mole$ of $P$ and $2 \, mole$ of $Q$ would be .......... $torr$.

MediumAIPMT 2005

View Solution

$A$ solution contains a $1:4$ molar ratio of pentane and hexane. At $20^{\circ}C$,the vapor pressure of pure pentane is $440 \ mm \ Hg$ and that of hexane is $120 \ mm \ Hg$. What is the mole fraction of pentane in the vapor phase?

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature. The mole fraction of toluene in the vapour phase,which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$,will be:

Similar Questions

The relative lowering in vapour pressure of an aqueous solution of urea which is $6 \%$ by weight,is

The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm \ Hg$ and $700 \ mm \ Hg$ respectively at $350 \ K$. Find out the composition of the liquid mixture if total vapour pressure is $600 \ mm \ Hg$. Also,find the composition of the vapour phase.

An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water,the rate at which water molecules :

The vapour pressure of two liquids $P$ and $Q$ are $80$ and $60 \, torr$,respectively. The total vapour pressure of the solution obtained by mixing $3 \, mole$ of $P$ and $2 \, mole$ of $Q$ would be .......... $torr$.

$A$ solution contains a $1:4$ molar ratio of pentane and hexane. At $20^{\circ}C$,the vapor pressure of pure pentane is $440 \ mm \ Hg$ and that of hexane is $120 \ mm \ Hg$. What is the mole fraction of pentane in the vapor phase?

Vedclass Test Series

Exam Paper Generator

Online Exam Module