Question

\( 2.5 \, \text{g} \) of a non-volatile, non-electrolyte is dissolved in \( 100 \, \text{g} \) of water at \( 25^\circ \text{C} \). The solution showed a boiling point elevation by \( 2^\circ \text{C} \). Assuming the solute concentration is negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is ______ mm of Hg (nearest integer).
(Given: Molal boiling point elevation constant of water (\( K_b \)) = \( 0.52 \, \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \),1 atm pressure = \( 760 \, \text{mm of Hg} \), molar mass of water = \( 18 \, \text{g mol}^{-1} \))

Answer 1

Correct Answer: 707

Determine the Molality of the Solution:

The boiling point elevation \( \Delta T_b \) is related to molality (\( m \)) as follows: \[ \Delta T_b = K_b \times m \]

Given:

\[ \Delta T_b = 2^\circ C, \quad K_b = 0.52 \, \text{K kg mol}^{-1} \] \[ m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} \approx 3.85 \, \text{mol kg}^{-1} \]

Calculate the Moles of Solute:

Since molality \( m \) is defined as moles of solute per kilogram of solvent: \[ \text{moles of solute} = m \times \text{mass of solvent (in kg)} \] Given that the mass of solvent (water) is 100 g or 0.1 kg: \[ \text{moles of solute} = 3.85 \times 0.1 = 0.385 \, \text{moles} \]

Determine the Molar Mass of the Solute:

Given mass of solute = 2.5 g, \[ \text{Molar mass of solute} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{2.5}{0.385} \approx 6.49 \, \text{g/mol} \]

Calculate the Vapour Pressure Lowering:

The vapour pressure lowering \( \Delta P \) is given by: \[ \Delta P = P^0 \times \frac{\text{moles of solute}}{\text{moles of solvent}} \] where \( P^0 = 760 \, \text{mm Hg} \) and moles of solvent (water) = \[ \frac{100}{18} \approx 5.56 \, \text{moles}. \]

Calculate \( \Delta P \):

\[ \Delta P = 760 \times \frac{0.385}{5.56} \approx 52.61 \, \text{mm Hg} \]

Calculate the Vapour Pressure of the Solution:

\[ P_{\text{solution}} = P^0 - \Delta P = 760 - 52.61 \approx 707 \, \text{mm Hg} \]

Conclusion:

The vapour pressure of the resulting aqueous solution is approximately \( 707 \, \text{mm Hg} \).

Answer 2

Step 1: Given data.
- Mass of solute \( = 2.5 \, \text{g} \)
- Mass of solvent (water) \( = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Boiling point elevation \( \Delta T_b = 2^\circ \text{C} \)
- Boiling point elevation constant \( K_b = 0.52 \, \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \)
- Normal atmospheric pressure \( = 760 \, \text{mm Hg} \)
- Molar mass of water \( = 18 \, \text{g mol}^{-1} \)

Step 2: Calculate molality.
\[ m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} \approx 3.85 \, \text{mol/kg} \]

Step 3: Moles of solute.
\[ \text{moles solute} = m \times \text{kg solvent} = 3.85 \times 0.1 = 0.385 \, \text{mol} \]

Step 4: Molar mass of solute.
\[ M = \frac{\text{mass}}{\text{moles}} = \frac{2.5}{0.385} \approx 6.49 \, \text{g/mol} \]

Step 5: Mole fractions.
- Moles of solvent:
\[ \frac{100}{18} = 5.56 \, \text{mol} \]
- Mole fraction of solute:
\[ X_{\text{solute}} = \frac{0.385}{5.56 + 0.385} \approx 0.0649 \]
- Mole fraction of solvent:
\[ X_{\text{solvent}} = 1 - 0.0649 = 0.9351 \]

Step 6: Calculate vapor pressure of solution using Raoult’s law.
\[ P_{\text{solution}} = X_{\text{solvent}} \times P_0 = 0.9351 \times 760 \approx 711 \, \text{mm Hg} \]

Approximate and round to nearest integer:
\[ \boxed{707} \]