Question

Two liquids A and B form an ideal solution. At $300$K the vapor pressure of a solution containing $1$ mole of A and $3$ mole of B is $550$ mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapor pressure of solution increases by $10$ mm of Hg. Determine the vapour pressure of A and B in their pure states.

Answer 1

We can observe that there are two factors in this question: vapor pressure and the number of moles, by applying Raoult’s law to above question we can find out the vapor pressure of both the liquids. When the liquid evaporates, vapours are formed and the pressure exerted by the vapours is known as vapor pressure and it is highly affected by change in the temperature.

Complete step by step answer:
We can clearly observe in the question, we are given two liquids that form an ideal solution. At a standard temperature, under different pressure conditions and with adding different moles of the solutions A and B we can observe different results. To evaluate the vapour pressure, we will use Raoult’s law equation.
We know that Raoult’s law is the vapor pressure of a component at a given temperature is equal to the product of mole fraction and vapor pressure of that component. Raoult’s law is applied when the two liquid phases are pure or a combination of a mixture of substances that are the same that is homogeneous.
Hence,
${{\text{P}}_{{\text{solution }}}}{\text{ = }}{{\text{X}}_{{\text{solvent}}}}{\text{.}}{{\text{P}}_{{\text{solvent}}}}$
In this question there are two liquids A and B, so the equation can be
${{\text{P}}_{\text{A}}}{\text{ = }}{{\text{X}}_{\text{A}}}{{\text{P}}_{\text{A}}}$……………...For liquid A
${{\text{P}}_{{\text{B = }}}}{{\text{X}}_{\text{B}}}{{\text{P}}_{\text{B}}}$………………For liquid B
${{\text{P}}_{{\text{solution}}}}{\text{ = }}{{\text{P}}_{\text{A}}}{\text{ + }}{{\text{P}}_{\text{B}}}$
${\text{550 = }}{{\text{X}}_{\text{A}}}{{\text{P}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{{\text{P}}_{\text{B}}}$
${\text{550 = }}\dfrac{{\text{1}}}{{\text{4}}}{{\text{P}}_{\text{A}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{4}}}{{\text{P}}_{\text{B}}}{\text{ }}$
Also, $\left[ {\dfrac{{\text{1}}}{{\text{4}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{1}}}}}{{{{\text{n}}_{\text{1}}}{\text{ + n}}}}{\text{ and }}\dfrac{{\text{3}}}{{\text{4}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{2}}}}}{{{{\text{n}}_{\text{2}}}{\text{ + n}}}}} \right]$
On multiplying by $4$on both the sides we get,
${\text{2200 = }}{{\text{P}}_{\text{A}}}{\text{ + 3}}{{\text{P}}_{\text{B}}}$……………equation $1$
Now, $1$mole of B is added;
${{\text{X}}_{{\text{A }}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{1 + 4}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{5}}}$
${{\text{X}}_{\text{B}}}{\text{ = }}\dfrac{{\text{4}}}{{\text{5}}}$
And pressure is increased by$10$mmHg, so other equation will be,
${\text{560 = }}{{\text{X}}_{\text{A}}}{{\text{P}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{{\text{P}}_{\text{B}}}$
${\text{560 = }}\dfrac{{\text{1}}}{{\text{5}}}{{\text{P}}_{\text{A}}}{\text{ + }}\dfrac{{\text{4}}}{{\text{5}}}{{\text{P}}_{\text{B}}}$
On multiplying both the sides by $5$ we get,
${\text{2800 = }}{{\text{P}}_{\text{A}}}{\text{ + }}{{\text{P}}_{\text{B}}}$……………equation $2$
Subtract equation $1$ from equation $2$ ,
$4{P_{_B}} - 3{P_B}$
$= 2800 - 2200$
${{\text{P}}_{\text{B}}}{\text{ = 600 mmHg}}$
Substituting the value of $P_B$ in equation $1$, we get;
${{{P}}_{{A}}}{{ + 3 \times 600 = 2200}}$
${{\text{P}}_{{\text{A }}}} = 2200 - 1800$
${{\text{P}}_{{\text{A }}}}{\text{ = 400 mmHg}}$
So, the vapor pressure of A is obtained as $400$ mmHg and vapor pressure of B obtained as $600$ mmHg in their pure states.

Therefore, the option B is correct.

Note: We should remember that if there are two liquids which are homogeneous then they will form the binary mixture and the solution is called an ideal solution. But it must be noted that all the mixtures do not follow Raoult’s law, some show deviation also, for e.g., mixture of chloroform and acetone. It is assumed that the interparticle forces between two different molecules is equal to the interparticle forces between the same molecules which forms the basis of Raoult’s law.