Question

Three solution are prepared by adding ′w′ gm of 'A' into 1kg of water, 'w' gm of 'B' into another 1kg of water and w gm of 'C' in another 1kg of water (A, B, C are non electrolytic). Dry air is passed from these solutions in sequence ${\text{ A}} \to {\text{B}} \to {\text{C }}$.The loss in weight of solution A was found to be 2gm while solution B gained $0.5{\text{ gm}}$ and solution C lost 1gm. Then the relation between molar masses of A, B and C is:
A.${\text{}}{{\text{M}}_{\text{A}}}{\text{:}}{{\text{M}}_{\text{B}}}{\text{:}}{{\text{M}}_{\text{C}}} = 4:3:5{\text{ }}$
B.${\text{}}{{\text{M}}_{\text{A}}}{\text{:}}{{\text{M}}_{\text{B}}}{\text{:}}{{\text{M}}_{\text{C}}} = \dfrac{1}{4}:\dfrac{1}{3}:\dfrac{1}{5}{\text{ }}$
C.${\text{ }}{{\text{M}}_{\text{C}}}{\text{ > }}{{\text{M}}_{\text{A}}}{\text{ > }}{{\text{M}}_{\text{B}}}{\text{ }}$
D.${\text{ }}{{\text{M}}_{\text{B}}}{\text{ > }}{{\text{M}}_{\text{A}}}{\text{ > }}{{\text{M}}_{\text{C}}}{\text{ }}$

Answer 1

Using relative lowering of vapour pressure first we will calculate the vapour pressure of dry air and then using Dalton law we will get the relation between the molar mass and the partial pressure by simplifying the Dalton law equation.

Complete step by step answer:
It is given to us that,
Mass of ${\text{ A}} = {\text{w g }}$
Mass of ${\text{ B}} = {\text{w g }}$
Mass of ${\text{ C}} = {\text{w g }}$
Mass of water used for each solution is 1 Kg.
Now we know the vapour pressure of dry gas is zero, so when dry air is passed through the solution A, the solution A loses weight that means some vapours will get absorbed by the dry air and hence vapour pressure of Dry air will increase. The loss in mass will give us the relative lowering of vapour pressure.
We will get the Vapour pressure of dry air after passing through A solution
${{\text{P}}_{\text{A}}} = 2$
Now, the same air passes through B solution & B solution gains weight, hence we will get:
${{\text{P}}_{\text{B}}} = 2 - 0.5 = 1.5$
Similarly for the solutions C we will get:
${{\text{P}}_{\text{C}}} = 1.5 + 1 = 2.5$
According to the Dalton law of partial pressure the partial pressure is directly proportional to the mole fraction of that component. According to this the partial pressure the partial pressure of dry gas passed through each solution will be directly proportional to the mole fraction of the solute present. Hence,
${{\text{P}}_{\text{A}}} \propto {\chi _{\text{A}}}$ mole fraction in the number of moles of solute divided by total number of moles. Total number of moles is a constant and hence partial pressure is directly proportional to the number of moles.
${{\text{P}}_{\text{A}}} \propto {{\text{n}}_{\text{A}}}$ Number of moles is mass of solute divided by its molar mass. We will get the final relation as:
${{\text{P}}_{\text{A}}} \propto \dfrac{{{{\text{w}}_{\text{A}}}}}{{{{\text{M}}_{\text{A}}}}}$
Similarly for B and C:
${{\text{P}}_{\text{B}}} \propto \dfrac{{{{\text{w}}_{\text{B}}}}}{{{{\text{M}}_{\text{B}}}}}$ And ${{\text{P}}_{\text{C}}} \propto \dfrac{{{{\text{w}}_{\text{C}}}}}{{{{\text{M}}_{\text{C}}}}}$
We can write them as ratio:
${{\text{P}}_{\text{A}}}{\text{:}}{{\text{P}}_{\text{B}}}{\text{:}}{{\text{P}}_{\text{C}}} = \dfrac{{{{\text{w}}_{\text{A}}}}}{{{{\text{M}}_{\text{A}}}}}:\dfrac{{{{\text{w}}_{\text{B}}}}}{{{{\text{M}}_{\text{B}}}}}:\dfrac{{{{\text{w}}_{\text{C}}}}}{{{{\text{M}}_{\text{C}}}}}$
Since the mass is same for all three of them:
${\text{2:1}}{\text{.5:2}}{\text{.5}} = \dfrac{{\text{1}}}{{{{\text{M}}_{\text{A}}}}}:\dfrac{{\text{1}}}{{{{\text{M}}_{\text{B}}}}}:\dfrac{{\text{1}}}{{{{\text{M}}_{\text{C}}}}}$
Reversing the equation we will get:
${{\text{M}}_{\text{A}}}{\text{:}}{{\text{M}}_{\text{B}}}{\text{:}}{{\text{M}}_{\text{C}}} = \dfrac{{\text{1}}}{2}:\dfrac{{\text{1}}}{{1.5}}:\dfrac{{\text{1}}}{{2.5}}$
We can multiply it with $\dfrac{{\text{1}}}{2}$ to make it a whole number as:
${{\text{M}}_{\text{A}}}{\text{:}}{{\text{M}}_{\text{B}}}{\text{:}}{{\text{M}}_{\text{C}}} = \dfrac{{\text{1}}}{4}:\dfrac{{\text{1}}}{3}:\dfrac{{\text{1}}}{5}$

Hence the correct option is option B.

Note:
The above question is based on the concept of the colligative property that is relative lowering in vapour pressure. Colligative properties are those properties that depend upon the amount of solute present in the solution. That is why change in vapour pressure is equal to the change in weight or moles of substance.