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Question: If \({P_o}\) and \({P_s}\) are the vapour pressures of the solvent and the solution respectively , \...

If Po{P_o} and Ps{P_s} are the vapour pressures of the solvent and the solution respectively , n1{n_1} and n2{n_2} , are the mole fractions of the solvent and solute respectively, then:
A. Ps=Pon1{P_s} = {P_o}{n_1}
B. Ps=Pon2{P_s} = {P_o}{n_2}
C. Po=Psn2{P_o} = {P_s}{n_2}
D. Ps=Po(n1n2){P_s} = {P_o}(\dfrac{{{n_1}}}{{{n_2}}})

Explanation

Solution

Hint : Vapour pressure is a colligative property, the vapour pressure of a solution is directly proportional to the volume of solute present. Since less liquid molecules are present at the surface of the solution when a solute is present in a solvent, the vapour pressure decreases.
A mole fraction is a calculation that compares the number of moles in solution. It is calculated by dividing the number of moles of solutes by the total number of moles. A solution's mole fractions must add up to one.

Complete Step By Step Answer:
According to the question,
The vapour pressures of the solvent and the solution are Po{P_o} and Ps{P_s} respectively.
The mole fractions of the solvent and solute are n1{n_1} and n2{n_2} .
The relation between the four variables can be found out using Raoult's Law.
The Raoult’s law states that the vapour pressure of a solvent above a solution is equal to the vapour pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent existing.
The above law can be mathematically represented as:
Po−PsPo=nn+N\dfrac{{{P_o} - {P_s}}}{{{P_o}}} = \dfrac{n}{{n + N}}
nnis the amount of a constituent measured in moles and n+Nn + Nis the cumulative number of constituents measured in moles.
From the question,
⇒Po−PsPo=n2\Rightarrow \dfrac{{{P_o} - {P_s}}}{{{P_o}}} = {n_2}
⇒1−PsPo=n2\Rightarrow 1 - \dfrac{{{P_s}}}{{{P_o}}} = {n_2}
On Simplifying, we get:
⇒PsPo=1−n2\Rightarrow \dfrac{{{P_s}}}{{{P_o}}} = 1 - {n_2}
We know that 1−n2=n11 - {n_2} = {n_1}because a solution's mole fractions should add up to one.
So, we get:
⇒PsPo=n1\Rightarrow \dfrac{{{P_s}}}{{{P_o}}} = {n_1}
⇒Ps=n1×Po\Rightarrow {P_s} = {n_1} \times {P_o}
So, the answer is option AA

Note :
It must be noted that for ideal liquids, Raoult's law can be shown to be true. Raoult's law is based on the premise that both the liquid and vapour phases behave optimally. This implies that there is no contact between the two molecules in the liquid and the vapour.