Question

1 molal aqueous solution of an electrolyte \( A_2B_3 \) is 60% ionized. Calculate the boiling point of the solution.

Hint:

- The van’t Hoff factor (\( i \)) accounts for the number of particles formed in solution. - A higher \( i \) results in a greater elevation in boiling point. - Electrolytes with higher ionization lead to larger \( \Delta T_b \).

Answer 1

\( T_b = 374.768 \) K (or 374.918 K if considering the boiling point of water as 373.15 K) 
Step 1: Boiling Point Elevation Formula The boiling point elevation is given by: \[ \Delta T_b = i K_b m \] where, \( K_b = 0.52 \, K \cdot kg \cdot mol^{-1} \) (for water), \( m = 1 \) molal, \( i \) = van’t Hoff factor. 

Step 2: Calculation of Van't Hoff Factor The dissociation of \( A_2B_3 \) is: \[ A_2B_3 \rightarrow 2A^{+} + 3B^{-} \] Total particles before dissociation = 1, Total particles after dissociation = 5. Degree of ionization \( \alpha \) is given as 60\% (0.6). \[ i = 1 + \alpha (n - 1) \] \[ i = 1 + 0.6 (5 - 1) \] \[ i = 1 + 2.4 = 3.4 \] 

Step 3: Calculate Boiling Point Elevation \[ \Delta T_b = 3.4 \times 0.52 \times 1 \] \[ \Delta T_b = 1.768 \text{ K} \] \[ T_b = 373 + 1.768 = 374.768 \text{ K} \] If the boiling point of water is taken as 373.15 K, \[ T_b = 373.15 + 1.768 = 374.918 \text{ K} \]