Question

The vapor 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 solution is \[0.2\], What would be the mole fraction of solvent if the decrease in vapour pressure is \[20\] mm:
A. \[0.8\]
B. \[0.6\]
C. \[0.4\]
D. \[0.2\]

Answer 1

Hint: We know that, the partial pressure of the volatile component or gas is proportional to the mole fraction in the solution Partial pressure and vapour pressure and their forthcoming alteration because of changing the parameter have been basic points for the regeneration in the field. Raoult’s law gives us a quantitative relation between the partial vapour pressures of components present in the vapour solution phase.

Complete step by step solution
Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2. The French chemist, Francois Marta Raoult (1886) gave the quantitative relationship between them. The relationship is known as the Raoult’s law which states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
According to the question, the initial pressure is \[10\;mm\], final pressure is \[20\;mm\], mole fraction of solute is \[0.2\]. And we have to find the value of mole fraction of solvent.
We know that, for a very dilute solution, pressure is proportional to the mole fraction and this statement is given by Raoult’s law.
\[\dfrac{{\Delta P}}{{{P^0}}} = \dfrac{{{\chi _1}}}{{{\chi _2}}}\]
Where \[\Delta P\] is the initial pressure, \[{\chi _1}\] is the mole fraction of the solute and \[{\chi _2}\] is the mole fraction of the solvent. According to Raoult’s law as mentioned in the question, the initial pressure is $10\;mm$ and final pressure is $20\;mm$. Mole fraction is $0.2$. putting this value in equation
\[
{P^0} - {P_S} = {P^0} \times mole\;fraction\;of\;solute...................................(a)\\
10 = {P^0} \times 0.2...............................(1)\\
20 = {P^0} \times {\chi _2}...................................(2)
\]
When we solve the equation \[1\;and\;2\] we get, the value of ${\chi _2}$ is $0.4$.
We know that;
$
{\chi _1} + {\chi _2} = 1\\
{\chi _1} = 1 - {\chi _2}
$
\[
{\chi _1} = 1 - 0.4\\
 = 0.6
\]

Therefore, the correct answer is B.

Note:
Raoult’s law only works for ideal mixture. The partial vapor pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.