Question

Explain Raoult's Law. The vapour pressure of chloroform (\( \text{CHCl}_3 \)) and dichloromethane (\( \text{CH}_2\text{Cl}_2 \)) are 200 mm Hg and 4.5 mm Hg respectively at 298 K. Calculate the vapour pressure of the solution formed by mixing 51 g of \( \text{CHCl}_3 \) and 20 g of \( \text{CH}_2\text{Cl}_2 \) at 298 K.

Hint:

Raoult's law holds true for ideal solutions where the intermolecular forces between molecules of different components are similar. For non-ideal solutions, deviations from Raoult's law can occur.

Answer 1

Raoult's Law states that the partial vapor pressure of each volatile component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. Mathematically, it is represented as: \[ P_i = X_i P_i^{\circ} \] Where: - \( P_i \) is the partial vapor pressure of component \( i \),
- \( X_i \) is the mole fraction of component \( i \) in the solution,
- \( P_i^{\circ} \) is the vapor pressure of the pure component \( i \).
Given: - Vapor pressure of chloroform \( P_{\text{chloroform}}^{\circ} = 200 \, \text{mm Hg} \),
- Vapor pressure of dichloromethane \( P_{\text{dichloromethane}}^{\circ} = 4.5 \, \text{mm Hg} \),
- Mass of chloroform \( m_{\text{chloroform}} = 51 \, \text{g} \),
- Mass of dichloromethane \( m_{\text{dichloromethane}} = 20 \, \text{g} \),
- Molar mass of chloroform \( M_{\text{chloroform}} = 119.38 \, \text{g/mol} \),
- Molar mass of dichloromethane \( M_{\text{dichloromethane}} = 84.93 \, \text{g/mol} \).
Step 1: Calculate moles of each component.
For chloroform: \[ n_{\text{chloroform}} = \frac{51 \, \text{g}}{119.38 \, \text{g/mol}} = 0.427 \, \text{mol}
\] For dichloromethane: \[ n_{\text{dichloromethane}} = \frac{20 \, \text{g}}{84.93 \, \text{g/mol}} = 0.235 \, \text{mol}
\] Step 2: Calculate total moles in the solution.
Total moles = \( n_{\text{chloroform}} + n_{\text{dichloromethane}} = 0.427 + 0.235 = 0.662 \, \text{mol} \).
Step 3: Calculate mole fractions.
Mole fraction of chloroform: \[ X_{\text{chloroform}} = \frac{n_{\text{chloroform}}}{n_{\text{chloroform}} + n_{\text{dichloromethane}}} = \frac{0.427}{0.662} = 0.645
\] Mole fraction of dichloromethane: \[ X_{\text{dichloromethane}} = \frac{n_{\text{dichloromethane}}}{n_{\text{chloroform}} + n_{\text{dichloromethane}}} = \frac{0.235}{0.662} = 0.355
\] Step 4: Apply Raoult's Law to calculate the total vapor pressure.
Vapor pressure of the solution is the sum of the partial pressures: \[ P_{\text{total}} = X_{\text{chloroform}} P_{\text{chloroform}}^{\circ} + X_{\text{dichloromethane}} P_{\text{dichloromethane}}^{\circ}
\] \[ P_{\text{total}} = (0.645 \times 200) + (0.355 \times 4.5)
\] \[ P_{\text{total}} = 129 + 1.598 = 130.598 \, \text{mm Hg}
\] Thus, the vapor pressure of the solution is \( 130.6 \, \text{mm Hg} \).