Question

A current of dry air is passed through a solution of 2.64g of non-volatile solute in 30.0g of ether and then through a pure ether. The loss in weight of solution was 0.645g and of the ether 0.0345g, the molecular weight of the solid is:
A.122 g
B.12.2 g
C.244 g
D.135 g

Answer 1

Hint: This question is based on Raoult’s law and Colligative property i.e. relative lowering in vapour pressure.
Raoult’s law: In a solution partial vapour pressure of each component is directly proportional to mole fraction of that component.Increase in vapour pressure is directly proportional to decrease in the weight of solvent or solution.

Complete step by step answer:
Let us see the data given in the question
The mass of solute (${w_2}$) $ = $ 2.64 g
The mass of solvent i.e. ether (${w_1}$) $ = $ 30 g
Now, we know that the molar mass of solvent i.e. ether (${M_1}$) $ = $ 74.12 g/mole
Loss in weight of solution $ = $ 0.645 g
Loss in weight of solvent (ether) $ = $ 0.0345 g
Thus, according to the Raoult’s law
${P^0} - {P_s} \propto $ Loss in the weight of pure ether is 0.0345 g
${P_s} \propto $ Loss in the weight of solution is 0.645 g
Here, ${P^0}$ represents the vapour pressure of solvent
${P_s}$ represents the vapour pressure of solution
Now, the formula can be written as:
$\dfrac{{{P^0} - {P_s}}}{{{P_s}}} = \dfrac{{{w_2} \times {M_1}}}{{{w_1} \times {M_2}}}$
So, we will calculate the value of ${M_2}$ (molecular weight of the solid)
After substituting the values
$\dfrac{{0.0345}}{{0.645}} = \dfrac{{2.64 \times 74.12}}{{30 \times {M_2}}}$
\[ \Rightarrow {M_2} = \dfrac{{2.64 \times 74.12 \times 0.645}}{{30 \times 0.0345}}\]
\[ \Rightarrow {M_2} = 122g\]
Thus, the value of molecular mass of non-volatile solute or solid is 122 g
Hence, the correct option is A.

Additional Information:
Colligative properties are used to determine the molecular mass of non-volatile solute and the value of colligative properties depend only on the number of solute particles present per unit volume and not on its nature.
There are different colligative properties:
1. Elevation in boiling point.
2. Depression in freezing point
3. Lowering in vapour pressure
4. Osmotic pressure

Note:
When non-volatile solute is added in volatile solvent that changes the properties of solvent like vapour pressure, freezing point, osmotic pressure and boiling point. For a given solute-solvent ratio, molar mass is inversely proportional to colligative properties.