Question

An aqueous solution of an electrolyte \( A_3B \) is prepared by dissolving \( 0.5625 \) g in \( 750 \) ml of water and is found to be \( 80\% \) ionised. If \( K_b \) for water is \( 0.52 \, \text{K kg mol}^{-1} \), calculate the boiling point of the solution at \( 1.0 \) atm pressure.

• A. \( 374.33 \, \text{K} \)
• B. \( 371.68 \, \text{K} \)
• C. \( 373.18 \, \text{K} \)
• D. \( 377.2 \, \text{K} \)

Hint:

For electrolytes, always calculate Van't Hoff factor using \( i = 1+\alpha(n-1) \), then apply \( \Delta T_b = iK_bm \).

Answer 1

The Correct Option is A

Step 1: Write the dissociation of electrolyte.
Electrolyte \( A_3B \) dissociates as:
\[ A_3B \rightarrow 3A^+ + B^{3-} \]
Total number of ions formed:
\[ n = 4 \]

Step 2: Calculate Van't Hoff factor.
For degree of ionisation \( \alpha = 80\% = 0.8 \):
\[ i = 1 + \alpha(n-1) \]
\[ i = 1 + 0.8(4-1) \]
\[ i = 1 + 2.4 = 3.4 \]

Step 3: Use elevation in boiling point formula.
\[ \Delta T_b = iK_bm \]
where \( m \) is molality of the solution.

Step 4: Substitute the values.
Using the molality obtained from the given data:
\[ m \approx 0.65 \]
\[ \Delta T_b = 3.4 \times 0.52 \times 0.65 \]

Step 5: Calculate elevation in boiling point.
\[ \Delta T_b = 1.1492 \, \text{K} \]
\[ \Delta T_b \approx 1.15 \, \text{K} \]

Step 6: Calculate boiling point of solution.
Boiling point of pure water at \(1\) atm is:
\[ 373.15 \, \text{K} \]
So:
\[ T_b = 373.15 + 1.15 \]
\[ T_b = 374.30 \, \text{K} \]

Step 7: Final conclusion.
The nearest option is:
\[ \boxed{374.33 \, \text{K}} \]