Question

A dilute solution of \( K_2HgI_4 \) reagent is 95% ionised. What would be the approximate value of its van't Hoff factor?

• A. 1.85
• B. 1.50
• C. 2.05
• D. 2.90

Hint:

The van't Hoff factor accounts for the degree of ionisation of a solute in a solution. For incomplete dissociation, it is less than the theoretical maximum.

Answer 1

The Correct Option is D

Step 1: Understand the concept of van't Hoff factor.
The van't Hoff factor (\(i\)) is the number of particles into which a solute dissociates in solution. For a solution, if the solute is completely dissociated, then the van't Hoff factor \(i\) equals the number of particles produced. For incomplete dissociation, it can be less than this value.

Step 2: Ionisation of \( K_2HgI_4 \).
The dissociation of \( K_2HgI_4 \) can be represented as:
\[ K_2HgI_4 \rightarrow 2K^+ + HgI_4^{2-} \]
If 100% of the \( K_2HgI_4 \) dissociates, it would produce 3 ions: 2 \( K^+ \) and 1 \( HgI_4^{2-} \), so the ideal van't Hoff factor is \( i = 3 \).

Step 3: Use the ionisation percentage.
Since the solution is 95% ionised, the dissociation is 95% of the ideal value. Hence, the actual van't Hoff factor is:
\[ i = 3 \times 0.95 = 2.85 \]

Step 4: Approximate the van't Hoff factor.
The approximate value of the van't Hoff factor is \( i \approx 2.90 \), which corresponds to option (D).

Step 5: Conclusion.
Thus, the approximate value of the van't Hoff factor for the \( K_2HgI_4 \) solution is 2.90.