To find the freezing point of the solution, we will use the formula for freezing point depression, which is given by:

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

where:

• $$ \Delta T_f $$ is the freezing point depression.


• $$ i $$ is the van't Hoff factor (number of particles the solute splits into or forms in solution).


• $$ K_f $$ is the cryoscopic constant (freezing point depression constant), for water it's 1.86 K kg mol^-1.


• $$ m $$ is the molality of the solution.

Given:

• Solute is $$ K_3[Fe(CN)_6] $$, and molecular weight (M) = 329 g/mol.


• Mass of solute = 0.1 g.


• Mass of solvent (water) = 100 g.

First, we calculate the molality $$ m $$ of the solution:

$$ m = \frac{{\text{Moles of solute}}}{{\text{Kilograms of solvent}}} $$

Moles of solute:

$$ \text{Moles of solute} = \frac{{\text{Mass of solute}}}{{\text{Molecular weight}}} = \frac{0.1}{329} \text{mol} $$

$$ \text{Moles of solute} = 3.04 \times 10^{-4} \text{ mol} $$

Kilograms of solvent:

$$ \text{Kilograms of solvent} = 100 \text{ g} = 0.1 \text{ kg} $$

Now molality:

$$ m = \frac{3.04 \times 10^{-4} \text{ mol}}{0.1 \text{ kg}} = 0.00304 \text{ mol/kg} $$

Next, calculate the van't Hoff factor $$ i $$:

• The solute $$ K_3[Fe(CN)_6] $$ disassociates into 4 ions: 3 $$ K^+ $$ and 1 $$ [Fe(CN)_6]^{3-} $$. Therefore, $$ i = 4 $$.

Substituting these into the freezing point depression formula:

$$ \Delta T_f = i \cdot K_f \cdot m = 4 \cdot 1.86 \cdot 0.00304 $$

$$ \Delta T_f = 0.02259 \text{ K} $$

Convert to degrees Celsius and remember that this will lower the freezing point, so it should be negative (since the freezing point depression means the solution freezes at a lower temperature than pure solvent):

$$ \Delta T_f = -0.02259 \text{ °C} $$

So, $$ \Delta T_f = -2.259 \times 10^{-2} \text{ °C} $$

From the given options, the closest to this calculated value of $$ \Delta T_f $$ is:

Option A: -2.3 $$\times$$ 10^-2oC. Hence, Option A is the correct answer.