The freezing point (in°C) of a solution containing 0.1 g of ${\mathrm{K}}_3\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ (Mol. wt. 329) in 100 g ofwater (K_f = 1.86 K kg mol^-1) is

${\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$
Where m = Molality of the solution
(i.e.Number of moles of solute per 1000 g of the solvent)
Here $\mathrm{m}=\frac{0.1}{329}\times 10$
Thus ${\mathrm{\Delta T}}_{\mathrm{f}}=4\times 1.86\times \frac{0.1\times 10}{329}=2.3\times {10}^{-2}$
Thus ${\mathrm{T}}_{\mathrm{f}}=0-2.3\times {10}^{-2}=-2.3\times {{10}^{-2}}^{\circ }\mathrm{C}$