$\begin{array}{1 1} 1000\;g \\ 2040\;g \\ 1860\;g \\ 3042 \;g \end{array}$

Answer: 1000 g.

Depression of Freezing point, $\Delta\; T_f = i\;K_f\; m$, where $i$ is the van't Hoff factor, $K_f$ is the Freezing Point Depression Constant and $m$ is the molality.

Ethylene glycol does not dissociat; hence its van't Hoff factor $i= 1 $

Molality $m$ is the number of moles of solute per kg of solvent, $m = \large\frac{W}{M}$$ \times \large \frac{1000}{V \;mL}$

$\Rightarrow m = \large\frac{W}{62\times5}$$ = \large\frac{W}{310}$

Subsitituting in $\Delta\; T_f = i\;K_f\; m$, we get $6^{\circ} = 1.86 \times \large\frac{W}{310}$

$\Rightarrow W = \large\frac{6 \times 310}{1.86}$$ = 1000\;g$

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