Freezing Point Depression of Potassium Ferricyanide Solution

To find the freezing point of the solution, we use the formula for the depression in freezing point:

$\Delta T_f = i \cdot K_f \cdot m$

where:

• $i$ is the van 't Hoff factor,
• $K_f$ is the cryoscopic constant of water ($1.86\text{ K kg mol}^{-1}$),
• $m$ is the molality of the solution.

Step 1: Determine the van 't Hoff factor ($i$)
Potassium ferricyanide, $\text{K}_3[\text{Fe}(\text{CN})_6]$, is a strong electrolyte and undergoes complete dissociation in an aqueous solution:
$\text{K}_3[\text{Fe}(\text{CN})_6] \longrightarrow 3\text{K}^+ + [\text{Fe}(\text{CN})_6]^{3-}$

Since it produces $4$ ions per formula unit, the van 't Hoff factor is:
$i = 4$

Step 2: Calculate the molality ($m$) of the solution
Given:

• Mass of solute ($\text{K}_3[\text{Fe}(\text{CN})_6]$), $w = 0.1\text{ g}$
• Molar mass of solute, $M = 329\text{ g mol}^{-1}$
• Mass of solvent ($\text{H}_2\text{O}$), $W = 100\text{ g} = 0.1\text{ kg}$

The number of moles of solute is:
$n = \frac{w}{M} = \frac{0.1}{329}\text{ mol}$

The molality of the solution is:
$m = \frac{n}{W} = \frac{\frac{0.1}{329}\text{ mol}}{0.1\text{ kg}} = \frac{1}{329}\text{ mol kg}^{-1}$

Step 3: Calculate the depression in freezing point ($\Delta T_f$)
$\Delta T_f = 4 \times 1.86 \times \frac{1}{329} = \frac{7.44}{329} \approx 0.0226\text{ }^\circ\text{C} \approx 2.3 \times 10^{-2}\text{ }^\circ\text{C}$

Step 4: Determine the freezing point of the solution ($T_f$)
Since the normal freezing point of pure water is $T_f^\circ = 0^\circ\text{C}$:
$T_f = T_f^\circ - \Delta T_f = 0 - 2.3 \times 10^{-2}\text{ }^\circ\text{C} = -2.3 \times 10^{-2}\text{ }^\circ\text{C}$

Therefore, the correct option is (A).