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: Count Particles:
Start with 1 mol of BaCl$_2$. If a fraction $\alpha$ dissociates, we have $1-\alpha$ undissociated plus $\alpha$ of Ba$^{2+}$ and $2\alpha$ of Cl$^-$.

Step 2: Total:
Total moles $= 1-\alpha+\alpha+2\alpha = 1+2\alpha$. This is the factor $i$. Setting $1+2\alpha = 2.47$ gives $\alpha=0.735$.

Step 3: Answer:
73.5 percent, option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } 73.5\%} \]