A solution containing 62g ethylene glycol in 250g water is cooled...

Molecular weight of ethylene glycol

$\left({C}_{2}{H}_{6}{O}_{2}\right) = 62$

Moles of ethylene glycol $= \dfrac{Wt}{M . Wt}$

$= \dfrac{62}{62} = 1$

$∆ {T}_{f} = i \times {k}_{f} \times m$

$∴ ∆ {T}_{f} = 0 - \left(- 10\right) = 10$

$i = 1$ for ethylene glycol

$∆ {T}_{f} = 1 \times 86 \times \dfrac{1}{\dfrac{250}{1000}}$

$∆ {T}_{f} = 1.86 \times 4 = 7.44$

As $∆ {T}_{f} = 10$ it implies that some amount of water has frozen and this has led to a greater depression of freezing point.

$10 = 1.86 \times \dfrac{\dfrac{62}{62}}{\dfrac{Wtof{H}_{2}O}{1000}}$

${W}_{{H}_{2}O} = 186gm$

Hence, the weight of water that has frozen $\text{=}( 250 - 186 )gm$

$= 64gm$