Ethylene glycol is used as an antifreeze in cold climates. What mass of ethylene glycol should be added to 4 kg of water so that the solution freezes at minus 6 degree celsius? (Kf for water = 1.86 K kg mol minus1, molar mass of ethylene glycol = 62 g mol minus1)

Introduction / Context:
This numerical chemistry question involves colligative properties, specifically freezing point depression. Antifreeze solutions are classic real world applications of this concept. The learner must apply the formula that relates the lowering of freezing point to molality of the solute and then convert from moles of ethylene glycol to mass. Working through such problems strengthens understanding of solution behaviour and the practical use of colligative property constants.

Given Data / Assumptions:
- The solvent is water and its freezing point depression constant Kf is given as 1.86 K kg mol minus1.
- Desired freezing point of the solution is minus 6 degree celsius, so the freezing point depression is 6 K from zero degree for pure water.
- Mass of water used is 4 kg.
- Molar mass of ethylene glycol is 62 g mol minus1.
- We assume ethylene glycol is a non electrolyte and does not dissociate, so van hoff factor is 1.

Concept / Approach:
The key relation for freezing point depression is delta Tf equals Kf multiplied by m, where delta Tf is the decrease in freezing point and m is the molality of the solution. Molality is defined as moles of solute per kilogram of solvent. First we calculate the required molality using delta Tf and Kf. Then we compute the total moles of ethylene glycol required for 4 kg of water. Finally we convert this amount into grams using the molar mass. The nearest suitable value among the options is selected as the answer.

Step-by-Step Solution:
Step 1: Calculate the freezing point depression. The solution freezes at minus 6 degree celsius and pure water freezes at 0 degree celsius. Therefore, delta Tf = 6 K.
Step 2: Use the relation delta Tf = Kf multiplied by m to find molality m. Rearranging gives m = delta Tf divided by Kf.
Step 3: Substitute the values: m = 6 divided by 1.86, which gives approximately 3.2258 mol kg minus1. For simple calculation we can round this to about 3.23 mol kg minus1.
Step 4: Molality is moles of solute per kilogram of solvent. For 4 kg of water, moles of ethylene glycol required = m multiplied by 4, which is about 3.2258 multiplied by 4, giving around 12.9032 moles.
Step 5: Convert moles of ethylene glycol to mass using its molar mass 62 g mol minus1. Mass = 12.9032 multiplied by 62, which is very close to 800 g.
Step 6: Rounding to a suitable number of significant figures, we obtain a required mass of about 800.00 g of ethylene glycol.

Verification / Alternative check:
As a quick check, note that m equals 6 divided by 1.86 is about 3.2 mol kg minus1. For 4 kg solvent, that gives roughly 3.2 multiplied by 4 equals 12.8 moles. Multiplying 12.8 by 62 g mol minus1 gives 793.6 g, which is very close to 800 g. The small differences are due to rounding at intermediate steps. Therefore, a value of about 800 g is entirely consistent with the calculation and matches the option 800.00 g.

Why Other Options Are Wrong:
- 204.60 g: This mass would correspond to a much smaller molality and would not lower the freezing point by as much as 6 K.
- 304.30 g: This is still well below the roughly 800 g required, so the resulting solution would have a freezing point significantly higher than minus 6 degree celsius.

Why Other Options Are Wrong (continued):
- 400.00 g: This is about half of the required mass. Doubling this value leads to the correct approximate mass, showing that 400 g is insufficient to reach minus 6 degree celsius.

Common Pitfalls:
Students sometimes confuse molarity and molality or forget that the Kf formula uses molality based on kilograms of solvent, not solution. Another mistake is to use the mass of solution instead of the mass of water when calculating moles of solute. Rounding too early in the calculation can also cause noticeable discrepancies. It is best to carry at least three or four significant figures through the calculations and round only at the end. Finally, some learners may forget that ethylene glycol does not dissociate, so the van hoff factor remains 1.

Final Answer:
800.00 g of ethylene glycol should be added to 4 kg of water to lower the freezing point to approximately minus 6 degree celsius under the given conditions.