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Q1.15:
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Solution :

Step 1: Identifying Given Parameters and Theoretical Foundation

According to Raoult’s Law for a solution containing a non-volatile solute, the relative lowering of vapor pressure is equal to the mole fraction of the solute. The solvent is water, which has a normal boiling point of $373.15 \text{ K}$. At this temperature, the vapor pressure of pure water ($P_1^\circ$) is equal to the atmospheric pressure.

  • Pressure of solution ($P_s$) = $1.004 \text{ bar}$
  • Vapor pressure of pure water at boiling point ($P_1^\circ$) = $1.01325 \text{ bar}$
  • Mass of solute ($w_2$) = $2 \text{ g}$
  • Mass of solvent water ($w_1$) = $100 \text{ g} - 2 \text{ g} = 98 \text{ g}$
  • Molar mass of water ($M_1$) = $18.02 \text{ g/mol}$

Step 2: Formulating the Mathematical Expression

The relationship between vapor pressure lowering and molar mass is given by the formula [Per Raoult's Law]:

$\frac{P_1^\circ - P_s}{P_1^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$

Where $M_2$ is the molar mass of the non-volatile solute.

Step 3: Algebraic Substitution and Calculation

Rearranging the formula to solve for $M_2$:

$M_2 = \frac{w_2 \times M_1 \times P_1^\circ}{w_1 \times (P_1^\circ - P_s)}$

Variable Value
$w_2$ $2 \text{ g}$
$M_1$ $18.02 \text{ g/mol}$
$P_1^\circ$ $1.01325 \text{ bar}$
$w_1$ $98 \text{ g}$
$P_1^\circ - P_s$ $1.01325 - 1.004 = 0.00925 \text{ bar}$

Substituting the values into the equation:

$M_2 = \frac{2 \times 18.02 \times 1.01325}{98 \times 0.00925}$

$M_2 = \frac{36.5385}{0.9065}$

$M_2 \approx 40.31 \text{ g/mol}$

Visual Representation of Vapor Pressure Lowering

Solute Concentration Vapor Pressure P° (Pure Solvent) Ps (Solution)

Final Solution: The molar mass of the solute is 40.31 g/mol.


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