Question

The vapour pressure of acetone at ${20^ \circ }C$ is 185 torr. When 1.2g of a non-volatile substance was dissolved in 100g of acetone at ${20^ \circ }C$, its vapour pressure was 183 torr.
The molar mass (g/mol) of the substance is:
(A) 32
(B) 64
(C) 128
(D) 488

Answer 1

Hint: Here acetone will be the solvent and we need to find out the molecular weight of solute. We will use the concept and formula of lowering of vapour pressure according to which relative lowering of vapour pressure is equal to the mole fraction of the solute.
Formula used:
-Relative lowering of vapour pressure:
 $\dfrac{{{P^ \circ } - P}}{{{P^ \circ }}} = \dfrac{{{W_2} \times {M_1}}}{{{M_2} \times {W_1}}}$ (1)

Complete step by step solution:
-In the question we can see that the vapour pressure of acetone decreases from 185 torr to 183 torr on adding a non-volatile solute to it.
-We also know that the relative lowering of vapour pressure is equal to the mole fraction of the solute. Mathematically it can be written as:
$\dfrac{{{P^ \circ } - P}}{{{P^ \circ }}} = {X_2}$
Where, ${P^ \circ }$ = vapour pressure of solvent before adding solute to it
P = vapour pressure of solvent after adding solute to it
${X_2}$ = mole fraction of solute
= $\dfrac{{{n_2}}}{{{n_1} + {n_2}}}$ (for dilute solutions ${n_2} < < < {n_1}$)
= $\dfrac{{{n_2}}}{{{n_1}}}$ = $\dfrac{{{W_2} \times {M_1}}}{{{M_2} \times {W_1}}}$
Where, ${W_2}$ and ${M_2}$ = the given weight and molecular weights of solute and
${W_1}$ and ${M_1}$ = given weight and molecular weight of the solvent
So, this formula for lowering of vapour pressure can also be written as: (1)
$\dfrac{{{P^ \circ } - P}}{{{P^ \circ }}} = \dfrac{{{W_2} \times {M_1}}}{{{M_2} \times {W_1}}}$
-According to the question the solvent is acetone and we need to find out the molecular weight of the solute (${M_2}$). For this we will be using the equation (1) as given above.
The question has given us the following values:
${P^ \circ }$ = 185 torr
P = 183 torr
${W_1}$ = 100 g
${M_1}$ = 58 g (of acetone)
${W_2}$ = 1.2 g (of non-volatile solute)
${M_2}$ = ? (we need to calculate)
Putting these values in equation (1) we get:
$\dfrac{{185 - 183}}{{185}} = \dfrac{{1.2 \times 58}}{{{M_2} \times 100}}$
$\dfrac{2}{{185}} = \dfrac{{69.6}}{{{M_2} \times 100}}$
${M_2} = \dfrac{{69.6 \times 185}}{{100 \times 2}}$
= 64.38 g
Hence the molecular weight of the non-volatile solute is 64.38 g/mol.

The correct option will be: (B) 64

Note: Relative lowering of vapour pressure is a colligative property of solutions along with 3 others: depression in freezing point of the solvent, the elevation of the boiling point of the solvent and osmotic pressure of the solutions. These properties depend on the number of solute particles relative to the total number of particles present in solution (irrespective of its nature).