(પાણી માટે $K_f= 1.86\, K\, kg\, mol^{-1}$)
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 $=0^{\circ} C .$ )