Use freezing point depression formula

$\Delta T_f = i K_f m$

where:

• $i$ = van’t Hoff factor
• $K_f = 1.86\, \text{K kg mol}^{-1}$
• $m$ = molality

Find moles of solute

Given mass of $K_3[Fe(CN)_6] = 0.1\,\text{g}$ and molar mass $=329$

$n = \frac{0.1}{329} = 3.04 \times 10^{-4}\,\text{mol}$

Find molality

Mass of water $=100\,\text{g}=0.1\,\text{kg}$

$m = \frac{3.04\times 10^{-4}}{0.1} = 3.04 \times 10^{-3}\,\text{mol kg}^{-1}$

Find van’t Hoff factor

$K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$

Total particles $=4$, so

$i=4$

Calculate depression in freezing point

$\Delta T_f = 4 \times 1.86 \times 3.04 \times 10^{-3}$

$\Delta T_f \approx 2.26 \times 10^{-2}\,^\circ \text{C}$

So freezing point of solution is

$0 - 2.26\times 10^{-2} = -2.26\times 10^{-2}\,^\circ \text{C}$

Approx.

$-2.3 \times 10^{-2}\,^\circ \text{C}$

Option matching

This matches Option A.

Comparison with stored answer

Stored correct answer: A

Derived answer: A

Hence, they agree.