Question

1.00 molal aqueous solution of trichloroacetic acid is heated to its boiling point. The boiling point of this solution was found to be 100.18\(^\circ\)C. Calculate the Van’t Hoff factor for trichloroacetic acid. (Given: \( K_b \) for water = 0.512 K kg mol\(^{-1}\))

Hint:

If \( i<1 \) → Association of molecules If \( i>1 \) → Dissociation into ions

Answer 1

Step 1: Understanding the Formula for Boiling Point Elevation.
The change in boiling point (\(\Delta T_b\)) is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \] where:
\(\Delta T_b\) = Boiling point elevation (change in boiling point),
\(i\) = Van't Hoff factor (number of particles into which a solute dissociates),
\(K_b\) = Boiling point elevation constant of the solvent (water in this case),
\(m\) = Molality of the solution.

Step 2: Given Data.
- Boiling point of the solution = 100.18\(^\circ\)C
- Normal boiling point of water = 100\(^\circ\)C
- \(\Delta T_b = 100.18 - 100 = 0.18\) K
- Molality of the solution \(m = 1.00\) mol/kg
- \(K_b\) for water = 0.512 K kg mol\(^{-1}\)

Step 3: Rearranging the Formula to Solve for \(i\).
Rearranging the equation to find the Van’t Hoff factor \(i\):
\[ i = \frac{\Delta T_b}{K_b \cdot m} \] Substitute the known values:
\[ i = \frac{0.18}{0.512 \times 1.00} \] \[ i = \frac{0.18}{0.512} = 0.3516 \]

Step 4: Conclusion.
The Van’t Hoff factor (\(i\)) for trichloroacetic acid is approximately 0.352, which suggests that trichloroacetic acid does not fully dissociate in water.