Question

\( \text{K}_2\text{HgI}_4 \) is \( 40\% \) ionised in aqueous solution. The value of its van't Hoff factor (i) is:

• A. 1.6
• B. 1.8
• C. 2.2
• D. 2.0

Hint:

Remember that the complex ion \( [\text{HgI}_4]^{2-} \) remains intact as a single entity in aqueous solution; it does not dissociate further into mercury and iodine ions.

Answer 1

The Correct Option is B

Step 1: Write the dissociation.
The complex salt breaks as $\text{K}_2[\text{HgI}_4] \rightarrow 2\text{K}^+ + [\text{HgI}_4]^{2-}$.
Step 2: Count particles per formula unit.
Full dissociation gives $n = 3$ particles (two K$^+$ and one complex anion).
Step 3: Track particles using a small table idea.
Start with 1 mole. A fraction $\alpha$ dissociates, leaving $(1-\alpha)$ undissociated and producing $3\alpha$ ions.
Step 4: Add up the total particles.
\[ i = (1-\alpha) + 3\alpha = 1 + 2\alpha \]
Step 5: Insert $\alpha = 0.4$.
\[ i = 1 + 2(0.4) = 1 + 0.8 \]
Step 6: Finish.
\[ i = 1.8 \] This particle-counting route gives exactly the same value as the formula $i = 1+(n-1)\alpha$. \[ \boxed{i = 1.8} \]