1. Use freezing point depression formula

For a non-volatile, non-electrolyte solute:

$\Delta T_f = K_f m$

where:

• $\Delta T_f = 6\,\text{K}$ (freezing point lowered from $0^\circ C$ to $-6^\circ C$)
• $K_f = 1.86\,\text{K kg mol}^{-1}$
• $m$ = molality

So,

$m = \frac{\Delta T_f}{K_f} = \frac{6}{1.86} = 3.2258\,\text{mol kg}^{-1}$

2. Find moles of ethylene glycol needed

Mass of water $= 4\,\text{kg}$

Since

$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

therefore,

$\text{moles of solute} = m \times 4 = 3.2258 \times 4 = 12.9032\,\text{mol}$

3. Convert moles to mass

Molar mass of ethylene glycol $= 62\,\text{g mol}^{-1}$

$\text{mass} = 12.9032 \times 62 = 800\,\text{g}$

4. Match with options

Calculated value is

$\boxed{800\,\text{g}}$

Among the given options, the nearest/intended option is B if exact arithmetic were not considered? But option B is $400\,\text{g}$, which is far off. Option D is $804.32\,\text{g}$, which is close to the expected result only if one incorrectly uses solvent mass as $4.02\,\text{kg}$ or rounds inconsistently.

So the physically and mathematically correct value is $800\,\text{g}$, and the intended closest option appears to be D.

5. Option-wise conclusion

• A: 204.30 g — incorrect
• B: 400.00 g — incorrect
• C: 304.60 g — incorrect
• D: 804.32 g — closest/intended exam answer

Hence, for the given MCQ, the marked option would be:

$\boxed{\text{D}}$

But the exact calculation gives $800\,\text{g}$, not $804.32\,\text{g}$.