Question

An aqueous solution of $2\% $ non-volatile solute exerts a pressure of $1.004bar$ at the normal boiling point of the solvent. What is the molar mass of the solute?
a) $23.4gmo{l^{ - 1}}$
b) $41.35gmo{l^{ - 1}}$
c) $10gmo{l^{ - 1}}$
d) $20.8gmo{l^{ - 1}}$

Answer 1

Hint: Colligative properties are those properties of a system which depend only on the number of solute particles and not on the nature of solute. Relative lowering of vapour pressure is a colligative property that occurs when a non-volatile solute is added to a volatile solvent.
Formula used: $\dfrac{{P_A^\circ - P}}{{P_A^\circ }} = \dfrac{{{W_B} \times {M_A}}}{{{W_A} \times {M_B}}}$

Complete step by step answer:
Relative lowering of vapour pressure is a colligative property that occurs when a non-volatile solute is added to a volatile solvent. The escaping tendency of solvent decreases as some of the surface area is occupied by non-volatile solute particles. According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute particle.
This lowering can be calculated by the formula represented as:
$\dfrac{{P_A^\circ - P}}{{P_A^\circ }} = \dfrac{{{W_B} \times {M_A}}}{{{W_A} \times {M_B}}}$
Where the partial pressure of the pure solvent is $P_A^\circ $, $P$ is the partial pressure of solution formed, ${W_A}$ is the given mass of solvent, ${W_B}$ is the given mass of non-volatile solute, ${M_A}$ is the molar mass of solvent and ${M_B}$ is the molecular mass of non-volatile solute.
The solvent given here is water. The molar mass of water (${M_A}$) is determined as $18gmo{l^{ - 1}}$. The value of $P$ is $1.004bar$, the value of partial pressure of pure water is $1.013bar$.
The mass percentage of solute is equal to $2\% $. So the given mass of solute (${W_B}$) will be $2g$.
The given mass of solvent will be (${W_A}$)$100g - 2g = 98g$.
On substituting the values we get-
$\dfrac{{1.013bar - 1.004bar}}{{1.013bar}} = \dfrac{{2g \times 18gmo{l^{ - 1}}}}{{98g \times {M_B}}}$
On solving the equation we get,
\[
  {M_B} = \dfrac{{2 \times 18gmo{l^{ - 1}} \times 1.013}}{{98 \times 0.009}} \\
   \Rightarrow {M_B} = \dfrac{{36.468gmo{l^{ - 1}}}}{{0.882}} \\
   \Rightarrow {M_B} = 41.35gmo{l^{ - 1}} \\
\]
Hence the molar mass of non-volatile solute was $41.35gmo{l^{ - 1}}$.
The correct option is (b).

Note:
The other colligative properties like elevation in boiling point and depression in freezing point depend on the molality of the solution unlike relative lowering of partial pressure which mainly depends on the mole fraction of non-volatile solute according to Raoult’s law.