Increasing amount of solid `Hgl_(2)` is added to `1 L` of an aqueous solution containing `0.1 mol KI`. Which fo the following graphs do represent the variation of freezing point of the resulting with the amount of `Hgi_(2)` added?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze how the freezing point of the solution changes as we add increasing amounts of solid HgI2 to a solution of KI. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Solution
We start with a 1 L aqueous solution containing 0.1 mol of KI. KI is a strong electrolyte that dissociates completely in solution:
\[ \text{KI} \rightarrow \text{K}^+ + \text{I}^- \]
This means that for 0.1 mol of KI, we have 0.1 mol of K⁺ ions and 0.1 mol of I⁻ ions.

### Step 2: Determine the Effect of Adding HgI2
When we add solid HgI2 to the solution, it reacts with the iodide ions (I⁻) from KI to form the complex ion:
\[ \text{HgI}_2 + 2 \text{I}^- \rightarrow \text{HgI}_4^{2-} \]
This reaction consumes iodide ions and forms a complex, which effectively reduces the number of free I⁻ ions in the solution.

### Step 3: Calculate the Van't Hoff Factor (i)
The Van't Hoff factor (i) is a measure of the degree of dissociation of a solute in solution. For KI, since it dissociates into two ions, i = 2. However, as HgI2 is added and reacts with I⁻, the effective concentration of I⁻ decreases, which means that the effective i will be less than 2.

### Step 4: Freezing Point Depression
The freezing point depression (ΔTf) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( K_f \) is the freezing point depression constant,
- \( m \) is the molality of the solution.

As we add HgI2, initially, the freezing point will decrease because the effective concentration of ions increases due to the dissociation of KI. However, once all the I⁻ ions are consumed (when the reaction with HgI2 is complete), the addition of more HgI2 will not change the freezing point anymore since HgI2 does not dissociate further in the solution (it behaves as a non-electrolyte).

### Step 5: Graphical Representation
- Initially, as we add HgI2, the freezing point decreases.
- After all the I⁻ ions are consumed, the freezing point will stabilize and remain constant despite the addition of more HgI2.

### Conclusion
The graph that represents this behavior will show a decrease in freezing point initially, followed by a plateau where the freezing point remains constant as more HgI2 is added. The correct answer is option C.