Question

an ideal solution was found to have vapour pressure of 80 torr ,when molefraction of non volatile solute is 0.2 . what could be vapour pressure of pure solvent at same temperature?

Answer 1

mole fraction of solute = 0.2
mole fraction of solvent = 1 - 0.2 = 0.8
Psolution = Xsolvent*P°solvent
80 = 0.8*P°solvent
P°solvent = 100torr

Answer 2

Vapor pressure of pure solvent at same temperature is 100 torr.

Given:

The solution's vapor pressure = 80 torr

Solute's mole fraction = 0.2

To find:

Vapor pressure of pure solvent = ?

Solution:

We have to use the Raoult's law of vapor pressure. According to this, the product of mole fraction of solvent and "vapor pressure" of pure solvent is equal to the "vapor pressure" of solution.

It can be written as:

$P_{\text { solution }}=x_{\text {solvent}} \times P_{\text { solvent }}^{\circ}$

Where,

$P_{\text {solution}} = Solvent's \ vapor \ pressure$

$x_{\text { solvent }} = Solvent's \ mole \ fraction$

$P_{\text { solvent }}^{\circ} = Vapor \ pressure \ of \ the \ pure \ solvent$

$Mole \ fraction \ of \ solvent = 1 - 0.2 = 0.8$

$P_{\text { solvent }}^{\circ}=\frac{P_{\text {solution}}}{x_{\text {solvent}}}$

$=\frac{80}{0.8}$

$\therefore P_{\text { solvent }}^{\circ}=100 \ torr$