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Q1.19 (ii):
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: (ii) vapour pressure of water at 298 K.

Solution :

Step 1: Establishing the Theoretical Framework

To determine the vapor pressure of pure water ($P_1^\circ$), we utilize Raoult's Law for solutions containing a non-volatile solute. The relative lowering of vapor pressure is defined by the mole fraction of the solute ($x_2$):

$\frac{P_1^\circ - P_1}{P_1^\circ} = x_2 = \frac{n_2}{n_1 + n_2}$ [Per Raoult's Law]

Where:

  • $P_1^\circ$: Vapor pressure of pure water.
  • $P_1$: Vapor pressure of the solution.
  • $n_1$: Moles of solvent ($H_2O$, Molar mass = $18 \text{ g/mol}$).
  • $n_2$: Moles of solute (Molar mass = $M_2$).

Step 2: Formulating System Equations

Initial State (Solution A):

Mass of solute ($w_2$) = $30 \text{ g}$

Mass of solvent ($w_1$) = $90 \text{ g}$

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

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

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

Final State (Solution B after adding 18 g water):

New mass of solvent ($w_1'$) = $90 + 18 = 108 \text{ g}$

$n_1' = \frac{108}{18} = 6 \text{ mol}$

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

Step 3: Algebraic Simplification

Let $x = \frac{30}{M_2}$. Rewrite Equations 1 and 2:

Eq 1: $1 - \frac{2.8}{P_1^\circ} = \frac{x}{5+x} \implies \frac{2.8}{P_1^\circ} = 1 - \frac{x}{5+x} = \frac{5}{5+x} \implies 5+x = \frac{5P_1^\circ}{2.8}$

Eq 2: $1 - \frac{2.9}{P_1^\circ} = \frac{x}{6+x} \implies \frac{2.9}{P_1^\circ} = 1 - \frac{x}{6+x} = \frac{6}{6+x} \implies 6+x = \frac{6P_1^\circ}{2.9}$

Step 4: Solving for $P_1^\circ$

Subtracting the expression for $(5+x)$ from $(6+x)$:

$(6 + x) - (5 + x) = \frac{6P_1^\circ}{2.9} - \frac{5P_1^\circ}{2.8}$

$1 = P_1^\circ \left( \frac{6}{2.9} - \frac{5}{2.8} \right)$

$1 = P_1^\circ \left( \frac{16.8 - 14.5}{8.12} \right)$

$1 = P_1^\circ \left( \frac{2.3}{8.12} \right)$

$P_1^\circ = \frac{8.12}{2.3} \approx 3.53 \text{ kPa}$

Data Visualization of Vapor Pressure Relationship

Solvent Moles (n₁) Vapor Pressure (kPa) (5, 2.8) (6, 2.9)

Final Solution: The vapor pressure of water at 298 K is $3.53 \text{ kPa}$.


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