Question

At $20^\circ C$ the vapour pressure of diethyl ether is $442\;mm\;Hg$ When $6.4g$ of a non-volatile solute is dissolved in $50g$ of ether the vapour pressure falls to $410\;mm\;Hg$ The molecular weight of the solute is
(A) $\left( 1 \right)\;\;150$
(B) $\left( 2 \right)\;\;130.832$
(C) $\left( 3 \right)\;\;160$
(D) $\left( 4 \right)\;\;180$

Answer 1

Hint : We are talking about the vapour pressure here first we should know what vapour pressure is the pressure that is exerted by the vapour of a liquid on the surface of the liquid. We will calculate vapour pressure and using the information we will find the molecular weight of the solute.

Complete Step By Step Answer:
When we add a non-volatile solute to water or any other solvent it lowers the vapour pressure of that solvent. we will calculate the vapour pressure from the formula
${P_{solution}}\; = \;{P_{solvent\;}} \times \;{x_{solvent}}$
where ${P_{solution}}\;$ is the vapour pressure of the solution, $\;{P_{solvent\;}}$ is the vapour pressure of the solvent, and $\;{x_{solvent}}$ is the mole fraction of solvent. Formula of mole fraction is written as
${{\text{x}}_{solvent}}\; = \;\dfrac{{{{\text{n}}_{solvent}}}}{{{{\text{n}}_{solvent}}\; + \;{{\text{n}}_{solute}}}}$
where ${{\text{n}}_{solvent}}$ is the number of moles of solvent and ${{\text{n}}_{solute}}$ is the number of moles of solute present in the solution. Now let’s write the values given to us
Temperature of the solution, $T = 20^\circ C$
Vapour pressure of solvent (Diethyl ether), ${{\text{P}}_{solvent}}\; = \;442\;mm\;Hg$
Vapour pressure of the solution, ${{\text{P}}_{solution}}\; = \;410\;mm\;Hg$
weight of solute, ${{\text{w}}_1} = 6.4{\text{g}}$
weight of solvent (Diethyl ether), ${{\text{w}}_2} = 50{\text{g}}$
Molecular weight of solvent (Diethyl ether), ${{\text{M}}_2} = 74\;{\text{g}}$
Putting value in the vapour pressure formula we get,
$\dfrac{{410}}{{442}}\; = \;\;\;{{\text{x}}_{solvent}}$
now putting the value in the mole fraction formula, we get
$\dfrac{{410}}{{442}} = \dfrac{{\dfrac{{50}}{{74}}}}{{\dfrac{{50}}{{74}}\; + \dfrac{{6.4}}{{{M_1}}}}}$
where ${M_1}$ is the molecular mass of the solute. on further solving the above equation we get the value
${{\text{M}}_1}\; = \;130.832$
So, option $\left( 2 \right)$ is the correct answer.

Note :
Above question proves the fact that when a non—volatile solute is added to a solvent it lowers its vapour pressure. Carefully calculating the value of mole fraction of solute and solvent in the solution is necessary for the calculation of vapour pressure of the solution.