The quantities provided in the question are:

$K_f of water = 1.86 K mo{l}^{-1} Kg$

${\mathrm{\Delta T}}_{\mathrm{f}}=3.5$

We know the formula $\Delta {\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$

Substituting quantities provided in the question, we get:

$$\begin{gathered} 3.5=1.86\times \mathrm{m} \\ \Rightarrow \mathrm{m}=\frac{3.5}{1.86}=1.88 \end{gathered}$$

Additionally, $\mathrm{m}=\frac{\text{ mass of solute }}{\text{ molar mass of solute × mass of solvent }}$

Again, substituting:

$$\begin{gathered} 1.88=\frac{\text{ mass of solute (x)}}{\text{ 342 × 1 }} \\ \Rightarrow \mathrm{mass} \mathrm{of} \mathrm{solute} (x)=1.88\times 342 \\ \Rightarrow \mathrm{mass} \mathrm{of} \mathrm{solute} (x)=642.96 \end{gathered}$$

Hence, weight of separated ice $=1000-642.96\simeq 357g$

Therefore, the required answer is A) 357 g.