Question

The vapour pressure of benzene at a certain temperature is 640 mm Hg. A non-volatile and non-electrolytic solid, weighing 2.175 g, is added to 39.08 g of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular mass (in AMU) of the solid substance?
(A) 49.50
(B) 59.6
(C) 69.5
(D) 79.8

Answer 1

When we heat a liquid it changes into vapour. Vapour pressure is the pressure that is exerted by vapours above the liquid surface when both the states are in equilibrium. When we dissolve a non-volatile solute like glucose in the solvent the vapour pressure will be decreased since the contribution towards vapour pressure will only be due to the solvent and solute will occupy the surface and even hinder the solvent molecules to produce vapour.

Complete answer:
In order to solve this question we need to understand Rault’s law and calculate relative lowering of vapour pressure.
We have 2 components- the Non -volatile solute component denoted by $B$ and the volatile solvent component denoted by $A$.
As per the Rault’s law, in a solution the partial vapour pressure due to each volatile component will be directly proportional to the Mole fraction of that component.
$\Rightarrow$ ${P_A} \propto {X_A}$
$\Rightarrow$ ${P_A} = {P_A}^\circ {X_A} - - - (1)$
Where ${P_A}$ is the vapour pressure of the solution (i.e. after adding solute to solvent)
${P_A}^\circ$ is the vapour pressure of pure solvent $A$(i.e. in the absence of solute)
${X_A}$ is the mole fraction of component A,
$\Rightarrow$ ${X_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}} - - - (2)$
where ${n_A}$ and ${n_B}$ are the number of moles of $A$ and $B$ respectively.
Equation (1) can also be written as,
${P_A} = {P_A}^\circ (1 - {X_B}) - - - (3)$ Since ${X_A} + {X_B} = 1$
${P_A} = {P_A}^\circ - {P_A}^\circ {X_B} - - - (4)$
$\Rightarrow$ ${X_B} = \dfrac{{{P_A}^\circ - {P_A}}}{{{P_A}^\circ }} - - - - (5)$ [Relative lowering in vapour pressure]
${X_B}$ is the mole fraction of component B,
${X_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}} - - - (6)$
where ${n_A}$ and ${n_B}$ are the number of moles of $A$ and $B$ respectively.
Number of moles is calculated as given mass/Molar mass.
In the question we are given-
${P_A}^\circ$(Vapour pressure of pure solvent benzene)=640 mm Hg
Weight of solute=2.175g
Weight of benzene=39.08g
${P_A}$ (Vapour pressure of solvent benzene in presence of non –volatile non –electrolytic solid)=600 mmHg
Firstly we use Equation (6) to calculate the mole fraction of B component-
${X_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}$ here ${n_A}$ is much greater than ${n_B}$ since solvent is present in large quantities.
So the equation becomes- ${X_B} = \dfrac{{{n_B}}}{{{n_A}}}$
${n_B} = \dfrac{{2.175}}{M}$ and ${n_A} = \dfrac{{39.08}}{{78}}$(since Molar mass of benzene is 78 g/mole)
Hence,
$\Rightarrow$ ${X_B} = \dfrac{{\dfrac{{2.175}}{M}}}{{\dfrac{{39.08}}{{78}}}} = \dfrac{{\dfrac{{2.175}}{M}}}{{0.5}}$
Putting this value in Equation (5),
$\Rightarrow$ $\dfrac{{\dfrac{{2.175}}{M}}}{{0.5}} = \dfrac{{640 - 600}}{{640}}$
$\Rightarrow$ $\dfrac{{2.175}}{{0.5M}} = \dfrac{{40}}{{640}}$
$\Rightarrow$ $M = \dfrac{{2.175 \times 640}}{{0.5 \times 40}} = 69.5$
So option (C) is correct.

Note:
Vapour pressure cannot be calculated in an open container and always calculated in a closed container. The relative lowering of Vapour pressure is a well-known example of colligative property that depends only on the concentration or the amount of solute dissolved and not on its nature or identity or the type of chemical species present.