Question

The vapour pressure of a solvent is 80 mm. On
Sol adding a non-volatile solute its drops to 60 mm Hg.
The mole fraction of the solute in the solution is
​

Answer 1

$\checkmark$ Given:

The vapour pressure of a solvent is 80 mm. On  adding a non-volatile solute its drops to 60 mm Hg.

$\checkmark$ To Find:

The mole fraction of the solute in the solution is

​

$\checkmark$ Solution:

We know that,

$\red{\bigstar \ \boxed{\rm \chi=\dfrac{P^{0}-P^{s}}{P^{0}}}}$

Here,

$\rm P^{0}$ = vapour pressure of pure liquid

$\rm P^{s}$ = vapour pressure of non-volatile solute

According to the question,

We are asked to find the mole fraction of the solute in the solution

So, we must find "Χ"

Given that,

The vapour pressure of a solvent is 80 mm. On  adding a non-volatile solute its drops to 60 mm Hg

Hence,

• $\rm P^{0}$ = 0.80 atm
• $\rm P^{s}$ = 0.60 atm

Substituting the values,

We get,

$\blue{\implies \rm \chi=\dfrac{0.80-0.60}{0.8}}$

$\green{\implies \rm \chi=\dfrac{0.2}{0.8}}$

$\pink{\rm \implies \chi=0.25}$

Therefore,

The mole fraction of the solute in the solution = 0.25

Answer 2

$\huge\bold{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• The vapour pressure of a solvent is 80mm .

• When a non-volatile solute is added to that solvent, then it's vapour pressure falls to 60mm .

TO FIND :-

• The mole-fraction of the solute in that solution .

CALCULATION :-

$<font color=baby>$ Let,

✍️ “P°” be the vapour pressure of the solvent .

✍️ “$\rm\bold{P^s}$” be the vapour pressure of the non-volatile solute .

• P° = 80mm .

• $\rm\underline{P^s}$ = 60mm .

$\pink\bigstar\:\rm{\purple{\boxed{\red{\chi_{solute}\:=\:\dfrac{P^{\circ}\:-\:P^s}{P^{\circ}}\:}}}}$$<font color=green>$

$\rm{\implies\:\chi_{solute}\:=\:\dfrac{80\:-\:60}{80}\:}$

$\rm{\implies\:\chi_{solute}\:=\:\dfrac{20}{80}\:}$

$\rm{\implies\:\chi_{solute}\:=\:\dfrac{1}{4}\:}$

$\rm{\blue{\implies\:\chi_{solute}\:=\:0.25\:}}$

$\pink\therefore$ $<font color=purple>$ The mole fraction of the solute in the solution is “0.25” .