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Q1.19 (i):
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute

Solution :

Step 1: Establishing the Theoretical Framework

The relative lowering of vapour pressure is governed by Raoult’s Law for solutions containing a non-volatile solute. The relationship is expressed as:

$\frac{P_1^\circ - P_s}{P_1^\circ} = \frac{n_2}{n_1 + n_2}$

Where:

  • $P_1^\circ$: Vapour pressure of pure solvent ($\text{H}_2\text{O}$)
  • $P_s$: Vapour pressure of the solution
  • $n_2$: Moles of solute ($M_2$ is the molar mass)
  • $n_1$: Moles of solvent

[Per Raoult's Law of partial pressure in non-volatile solute systems]

Step 2: Defining Initial and Final Conditions

Initial State: Mass of solute ($w_2$) = $30\text{ g}$, Mass of solvent ($w_1$) = $90\text{ g}$, $P_s = 2.8\text{ kPa}$.

$n_1 = \frac{90}{18} = 5\text{ moles}$

$n_2 = \frac{30}{M_2}$

Equation 1: $\frac{P_1^\circ - 2.8}{P_1^\circ} = \frac{30/M_2}{5 + 30/M_2}$

Final State: Added $18\text{ g}$ of $\text{H}_2\text{O}$, so new $w_1 = 90 + 18 = 108\text{ g}$. $P_s = 2.9\text{ kPa}$.

$n_1 = \frac{108}{18} = 6\text{ moles}$

Equation 2: $\frac{P_1^\circ - 2.9}{P_1^\circ} = \frac{30/M_2}{6 + 30/M_2}$

Step 3: Algebraic Manipulation and Solving for $M_2$

Rearranging equations into the form $\frac{P_s}{P_1^\circ} = 1 - \frac{n_2}{n_1+n_2} = \frac{n_1}{n_1+n_2}$:

Eq 1: $\frac{2.8}{P_1^\circ} = \frac{5}{5 + 30/M_2} \implies P_1^\circ = \frac{2.8(5 + 30/M_2)}{5} = 2.8 + \frac{16.8}{M_2}$

Eq 2: $\frac{2.9}{P_1^\circ} = \frac{6}{6 + 30/M_2} \implies P_1^\circ = \frac{2.9(6 + 30/M_2)}{6} = 2.9 + \frac{14.5}{M_2}$

Equating the expressions for $P_1^\circ$:

$2.8 + \frac{16.8}{M_2} = 2.9 + \frac{14.5}{M_2}$

$\frac{16.8 - 14.5}{M_2} = 2.9 - 2.8$

$\frac{2.3}{M_2} = 0.1$

$M_2 = \frac{2.3}{0.1} = 23\text{ g mol}^{-1}$

Step 4: Visualizing the Vapour Pressure Gradient

Solvent Mass (g) Pressure (kPa) (90, 2.8) (108, 2.9)

Final Solution: The molar mass of the solute is 23 g mol⁻¹.


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