The vapor pressure of a glucose (C6H12O6) solution is 17.01 mmHg at 20 °C, while that of pure water is 17.25 mmHg at the same temperature. Calculate the molality of the solution.

Steps to Solve

1. Calculate the change in vapor pressure ($\Delta P$)

To find $\Delta P$, subtract the vapor pressure of the solution from the vapor pressure of pure water.

[ \Delta P = P_1^\circ - P = 17.25 , \text{mmHg} - 17.01 , \text{mmHg} = 0.24 , \text{mmHg} ]

2. Calculate the mole fraction of the solute ($X_2$)

Using the formula $X_2 = \frac{\Delta P}{P_1^\circ}$, we can find the mole fraction of glucose (the solute).

[ X_2 = \frac{0.24 , \text{mmHg}}{17.25 , \text{mmHg}} \approx 0.0139 ]

3. Calculate the mole fraction of the solvent ($X_1$)

Since the total mole fraction must equal 1, we can find the mole fraction of the solvent (water):

[ X_1 = 1 - X_2 = 1 - 0.0139 \approx 0.9861 ]

4. Calculate the molality ($m$)

Next, we know that molality ($m$) is calculated as:

[ m = \frac{X_2}{X_1} \times \frac{1000 , \text{g of solvent}}{molar , mass , of , solvent , (g/mol)} ]

The molar mass of water (H₂O) is approximately 18.015 g/mol, so:

[ m = \frac{0.0139}{0.9861} \times \frac{1000}{18.015} \approx 0.783 , \text{m} ]