Question

Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be ⋯⋯ respectively.
(A) 200 and 300
(B) 300 and 400
(C) 400 and 600
(D) 500 and 600

Answer 1

Hint: In order to find the vapour pressure of given components in pure states, we need to apply the Raoult's law. We are given the number of moles and from these values we can find the mole fraction of components and by substituting the given values in the Raoult's law and on comparing the conditions in the given question, we would be able to find the vapour pressure in pure states.

Complete step by step answer:
-Let's start with the concept of Raoult's law. It gives us the relationship between mole fraction and vapour pressure for a volatile liquid. According to Raoult's law, for a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to the component's mole fraction.
 - If there are two components present the Raoult's law can be represented as follows
\[{{P}_{total}}={{X}_{x}}{{P}_{x}}^{o}+{{X}_{y}}{{P}_{y}}^{o}\] ⋯⋯equation (1)
Where
    ${{P}_{total}}$ is the total vapour pressure of the ideal solution.
    ${{X}_{x}}$ is the mole fraction of component X.
    ${{P}_{x}}^{o}$ is the vapour pressure of X in pure state.
    ${{X}_{y}}$ is the mole fraction of component Y.
    ${{P}_{y}}^{o}$ is the vapour pressure of Y in pure state.

-We are asked to find the ${{P}_{x}}^{o}$ and ${{P}_{y}}^{o}$. Mole fraction of component X can be found by dividing the number of moles of X by the total number of moles of X and Y and similarly the mole fraction of Y can also be found.

-Let's take the first condition in which vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550mm Hg. From the equation (1) of Raoult's law we can write,
\[550=\left( \dfrac{1}{1+3} \right){{P}_{x}}^{o}+\dfrac{3}{4}{{P}_{y}}^{o}\] ⋯⋯equation (2)

On simplifying the above equation, we can write
\[550=\dfrac{{{P}_{x}}^{o}}{4}+\dfrac{3{{P}_{y}}^{o}}{4}\] ⋯⋯equation (3)
The above equation on rearrangement becomes
\[550\times 4={{P}_{x}}^{o}+3{{P}_{y}}^{o}\]
\[2200={{P}_{x}}^{o}+3{{P}_{y}}^{o}\] ⋯⋯equation (4)

 - It's given in the question as when 1 mol of Y is further added to this solution, vapour pressure of the solution is increased by 10 mm Hg. Hence the new total vapour pressure becomes 260 mmHg and the number of moles also changes. The equation of Raoult's law can be written as
\[560=\dfrac{1}{5}{{P}_{x}}^{o}+\dfrac{4}{5}{{P}_{y}}^{o}\] ⋯⋯equation (5)
The above equation on rearrangement becomes
\[560\times 5={{P}_{x}}^{o}+4{{P}_{y}}^{o}\]

\[2800={{P}_{x}}^{o}+4{{P}_{y}}^{o}\] ⋯⋯equation (6)
The equation (4) is subtracted from equation (6) and the following relation is obtained
\[{{P}_{y}}^{o}=600\]
 Thus, the vapour pressure of Y in pure state is obtained. Substituting this value in equation (4) we get
\[2200={{P}_{x}}^{o}+\left( 3\times 600 \right)\]
\[{{P}_{x}}^{o}=400\]
 Thus, the vapour pressure of X and Y in their pure states will be 400 mmHg and 600 mmHg respectively.
So, the correct answer is “Option C”.

Note: It should be noted that in real life the Raoult's law is not obeyed in the entire range of concentrations. Those solutions which obey Raoult's law in the entire range of concentrations are known as ideal solutions and most of the solutions show either positive or negative deviation from the ideal behaviour.