Question

The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of solute in a solution is 0.2, what would be the mole fraction of the solvent if the decrease in vapour pressure is 20 mm of Hg:
(A) 0.8
(B) 0.6
(C) 0.4
(D) 0.2

Answer 1

Hint: When a non-volatile solute is added to a pure solvent, relative lowering of vapour pressure takes place and this lowering is dependent on the mole fraction of components. From Raoult's law, we will be able to formulate this relation as $\dfrac{\Delta P}{{{P}^{o}}}={{X}_{2}}$ and from this equation we could find the mole fraction of solvent.

Complete step by step solution:
-Let's start with the concept of Raoult's law. It gives us the relationship between mole fraction and vapour pressure for a solution.
-In the given question the vapour of a solvent is being decreased on the addition of a non-volatile solute. Thus, the solution undergoes relative lowering of vapour pressure
-From Raoult's law for dilute solutions which contain non-volatile solutes, the mole fraction of solute is equal to the relative lowering of vapour pressure. This relation can be mathematically represented as follows
\[\dfrac{\Delta P}{{{P}^{o}}}={{X}_{2}}\]
Where $\dfrac{\Delta P}{{{P}^{o}}}$ is the relative lowering of vapour pressure
${ X }_{ 2 }$ is the mole fraction of the solute.
Let's write the initial condition as follows
The relative vapour pressure,$\Delta P_{ 1 }=10\quad mm\quad Hg$
Mole fraction of solute ${ X }_{ { B }_{ 1 } }$= 0.2
After adding the non-volatile solute, the values change into
The relative vapour pressure,$\Delta P_{ 2 }=20\quad mm\quad Hg$
We are supposed to find the mole fraction of solute ${ X }_{ { B }_{ 2 } }$ and mole fraction of the solvent ${ X }_{ { A }_{ 2 } }$
From the Raoult's law we could write as follows
\[\dfrac { { \Delta }P_{ 1 } }{ P^{ o } } ={ X }_{ { B }_{ 1 } }\]
\[\dfrac { { \Delta }P_{ 2 } }{ P^{ o } } ={ X }_{ { B }_{ 2 } }\]
On dividing the above equations, we get
\[\dfrac { { \Delta }P_{ 1 } }{ { \Delta }P_{ 2 } } =\dfrac { { X }_{ { B }_{ 1 } } }{ { X }_{ { B_{ 2 } } } }\]
On substituting the given values of relative vapour pressure and mole fraction we get
\[\dfrac { 10 }{ 20 } =\dfrac { { 0.2 } }{ { X }_{ { B_{ 2 } } } }\]
Therefore ${ X }_{ { B }_{ 2 } }$ $=0.2×2$
$=0.4$
We got the mole fraction of the solute and in order to find the mole fraction of solvent subtract the mole fraction of solute from 1, since the sum of mole fraction of solvent and solute will always be equal to one.
\[{ X }_{ { A_{ 2 } } }=1-{ X }_{ { B_{ 2 } } }\]
Thus, the mole fraction of solvent ${ X }_{ { A }_{ 2 } }$is
\[{ X }_{ { A_{ 2 } } }=1-0.4\]
\[{ X }_{ { A_{ 2 } } }=0.6\]

Therefore, the answer is option (B) 0.6.

Note: It should be noted that the vapour pressure of a pure solvent is caused by the number of molecules evaporating from the solvent's surface. On the addition of a non-volatile solute to the solvent, the escape of solvent particles from the surface reduces and as a result, the lowering of vapour pressure takes place.