$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \frac{\mathrm{w}}{\mathrm{m}}\times \frac{1000}{\mathrm{W}\left(\mathrm{In} \mathrm{gms}\right)}$

$10=1.86\times \frac{25}{62}\times \frac{1000}{W}$

$\mathrm{W}\left(\mathrm{weight} \mathrm{of} \mathrm{solvent}\right)= 75 \mathrm{gr}$

$\left[{\mathrm{w}}_{\mathrm{glycol} \mathrm{remains} \mathrm{constant} \mathrm{on} \mathrm{cooling}; \mathrm{only} \mathrm{water} \mathrm{freezes} }\right]$

$$\begin{gathered} \mathrm{Therefore}, \mathrm{Ice} \mathrm{separated} ={\mathrm{W}}_{{\mathrm{H}}_2\mathrm{o} \mathrm{initial} \mathrm{taken}}-{\mathrm{W}}_{{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{remaining}\right)}} \\ =100-75=25 \mathrm{gm} \end{gathered}$$