The freezing point of 0.05 m solutions of a non-electrolyte in water is

As we know, the depression in freezing point is 

${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{i}\times$molality 

From question, molal concentration $={\mathrm{C}}_{\mathrm{m}}=0.05\mathrm{m}$

For, non-electrolytes, $$\begin{gathered} \mathrm{i}=1 \\ {\mathrm{K}}_{\mathrm{f}}=1.86 \end{gathered}$$

$\therefore {\mathrm{\Delta T}}_{\mathrm{f}}=1.86\times 1\times 0.05=0{.0930}^{\mathrm{o}}\mathrm{C}$

hence, Freezing point of solution $=0-0{.093}^{\mathrm{o}}\mathrm{C}=-0{.093}^{\mathrm{o}}\mathrm{C}$