Question

van't Hoff factor for \(\text{BaCl}_2\) is \(2.47\), calculate the percentage dissociation of in its aqueous solution.

• A. \(25.6\%\)
• B. \(35.7\%\)
• C. \(73.5\%\)
• D. \(13.6\%\)

Hint:

For BaCl2, i = 1 + 2 alpha since each formula unit gives three ions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The van't Hoff factor relates to degree of dissociation. For \(\text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^-\), one formula unit gives \(n=3\) particles.

Step 2: Key Formula:
\[ i = 1 + (n-1)\alpha = 1 + 2\alpha \]

Step 3: Calculation:
\[ 2.47 = 1 + 2\alpha \Rightarrow \alpha = \frac{1.47}{2} = 0.735 \]
Percentage dissociation \(= 0.735 \times 100 = 73.5\%\).

Step 4: Check the Other Options:
25.6%, 35.7% and 13.6% would give i of 1.51, 1.71 and 1.27. None of these equals 2.47. So (C) is correct.

Final Answer:
BaCl\(_2\) is 73.5% dissociated, option (C). \[ \boxed{\text{(C) } 73.5\%} \]