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Q1.34:
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Solution :

Step 1: Identifying Given Parameters

The system involves a non-volatile solute (glucose) dissolved in a volatile solvent (water). We apply Raoult’s Law for solutions containing non-volatile solutes.

  • Vapour pressure of pure solvent ($P_1^\circ$) = $17.535 \text{ mm Hg}$
  • Mass of solute ($w_2$, glucose, $C_6H_{12}O_6$) = $25 \text{ g}$
  • Mass of solvent ($w_1$, water, $H_2O$) = $450 \text{ g}$
  • Temperature ($T$) = $293 \text{ K}$

Step 2: Determining Molar Masses

We calculate the molar masses ($M$) based on atomic masses ($C=12.01, H=1.008, O=16.00 \text{ g/mol}$):

  • Molar mass of Glucose ($M_2$) = $(6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.16 \text{ g/mol}$
  • Molar mass of Water ($M_1$) = $(2 \times 1.008) + 16.00 = 18.016 \text{ g/mol}$

Step 3: Calculating Number of Moles ($n$)

Using $n = \frac{\text{mass}}{\text{Molar Mass}}$:

  • Moles of Glucose ($n_2$) = $\frac{25 \text{ g}}{180.16 \text{ g/mol}} \approx 0.1388 \text{ mol}$
  • Moles of Water ($n_1$) = $\frac{450 \text{ g}}{18.016 \text{ g/mol}} \approx 24.9778 \text{ mol}$

Step 4: Applying Raoult's Law

Per Raoult's Law for non-volatile solutes, the relative lowering of vapour pressure is equal to the mole fraction of the solute ($x_2$):

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

Substituting the values:

$\frac{17.535 - P_1}{17.535} = \frac{0.1388}{24.9778 + 0.1388}$

$\frac{17.535 - P_1}{17.535} = \frac{0.1388}{25.1166} \approx 0.005526$

Step 5: Solving for $P_1$

Rearranging the equation to solve for $P_1$ (vapour pressure of the solution):

$17.535 - P_1 = 0.005526 \times 17.535$

$17.535 - P_1 = 0.0969$

$P_1 = 17.535 - 0.0969 = 17.4381 \text{ mm Hg}$

Visual Representation of Vapour Pressure Depression

Mole Fraction of Glucose ($x_2$) Vapour Pressure ($P$) $P_1^\circ$ Solvent + Solute

Final Solution: The vapour pressure of the glucose solution at 293 K is 17.438 mm Hg.


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