Question

Calculate the boiling point of a solution containing 0.61 g of benzoic acid (122 \( g mol^{-1} \)) in 5 g of \( CS_2 \) in which it dimerises to 88%. \( T_b^\circ(CS_2) = 46.2^\circ C, K_b = 2.3 K kg mol^{-1} \).

Hint:

Always check if the solute associates or dissociates. Dimerization (n=2) is common for organic acids in non-polar solvents like $CS_2$ or Benzene.

Answer 1

To calculate the boiling point elevation, we can use the formula:
\[ \Delta T_b = K_b \cdot m \] Where:
- \(\Delta T_b\) is the boiling point elevation.
- \(K_b\) is the ebullioscopic constant.
- \(m\) is the molality of the solution.
First, let's calculate the molality (m) of the solution, which is given by:
\[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} \] 1. Calculate the moles of benzoic acid:
The mass of benzoic acid = 0.61 g
Molar mass of benzoic acid = 122 g/mol
\[ \text{mol of benzoic acid} = \frac{0.61 \, \text{g}}{122 \, \text{g/mol}} = 0.005 \, \text{mol} \] 2. Calculate the effective number of moles of benzoic acid:
Given that 88% of the benzoic acid dimerizes, only 12% remains as monomeric benzoic acid.
Moles of monomeric benzoic acid = \( 0.12 \times 0.005 \, \text{mol} = 0.0006 \, \text{mol} \) 3. Calculate the molality (m):
The mass of \( CS_2 \) (solvent) = 5 g = 0.005 kg
\[ m = \frac{0.0006 \, \text{mol}}{0.005 \, \text{kg}} = 0.12 \, \text{mol/kg} \] 4. Calculate the boiling point elevation \(\Delta T_b\):
\[ \Delta T_b = K_b \times m = 2.3 \, \text{K kg/mol} \times 0.12 \, \text{mol/kg} = 0.276 \, \text{K} \] 5. Calculate the final boiling point:
The normal boiling point of \( CS_2 \) is \( 46.2^\circ C \). Therefore, the boiling point of the solution is:
\[ T_b = 46.2^\circ C + 0.276^\circ C = 46.476^\circ C \] Final Answer: The boiling point of the solution is approximately \( 46.48^\circ C \).