a
Depression in Freezing point \(\Delta T_{f}=\frac{W_{1} \times K_{f} \times 1000}{M_{1} \times W_{2}}\)

where \(\quad W_{1}=\) Weight of Solute

\(W_{2}=\) Weight of solvent

\(M_{1}=\) Molar mass of solute

\(K_{f}=\) Freezing point deprssion constant

Now, \(\Delta T_{f}=\begin{array}{cc}1.86 \times 68.5 \times 1000 \\ 342 \times 1000\end{array}=0.372 C\)

Now, \(\quad \Delta T_{f}=T^{\circ}-T_{f}\)

So, \(\quad T_{f}=0-0.372=-0.372 C\)

(Freezing point of purewater \(=0^{\circ} C .\) )