Question

At 88°C benzene has a vapour pressure of 900 torr and toulene has a vapour presssure of 360 torr.What is the mole fraction of benzene in the mixture with toulene that will boil at 88°C at 1atm pressure, benzene-toulene form an ideal solution.

Answer 1

Given:

At 88°C benzene has a vapour pressure of 900 torr and toulene has a vapour presssure of 360 torr

Benzene-toulene form an ideal solution

To Find :

Mole fraction of benzene in the mixture with toulene that will boil at 88°C at 1atm pressure.

Solution :

Benzene-toulene form an ideal solution

=> Solution will follow Roult's law

i.e. $P_{Total}$ = $P_{B} X_{B} + P_{T} X_{T}$___________(1)

where $P_{B}$ & $P_{T}$ are vapour pressure of pure Benzene & Toulene respectively.

$X_{B}$ & $X_{T}$ are mole fraction of Benzene & Toulene respectively.

Also,  + $X_{T}$ = 1

=>  $X_{T}$ = 1 - $X_{B}$___________(2)

Put (2) in (1)

$P_{Total}$ = $[P_{B} - P_{T} ] * X_{B} + P_{T}$

760 = [ 900 - 360 ]*$X_{B}$ + 360

400 = 540*$X_{B}$ + 360

400 = 540*$X_{B}$

$X_{B}$ = 400/540 = 0.74

Hence , Mole fraction of benzene is 0.74