Coolant by Molality: Meeting a Freezing-Point Specification
A college workshop prepares 1.50 kg of an aqueous coolant for a demonstration chiller. A student adds 250 g of ethylene glycol (molar mass 62.0 g mol^-1) and stirs until homogeneous. The label on the chiller says the coolant should remain liquid down to -8.0 °C and should not exceed 103 °C in normal use. The student claims the mixture is suitable because 'ethylene glycol lowers the freezing point and raises the boiling point.' Another student warns that the calculation should use molality, not molarity, and that the actual behaviour may differ from ideal. For water, Kf = 1.86 K kg mol^-1 and Kb = 0.512 K kg mol^-1. Assume ethylene glycol is non-volatile and does not dissociate. The density of the final solution is about 1.02 g mL^-1, and the laboratory is at 25 °C. The group must decide whether to use the coolant as prepared or modify the composition before the demonstration.
1. Calculate the expected freezing point and boiling point of the coolant. Does the mixture meet both label specifications? If not, how much extra ethylene glycol is needed to reach -8.0 °C, assuming the same 1.50 kg of water?
Read the worked answer
Moles of ethylene glycol = 250 g / 62.0 g mol^-1 = 4.032 mol.
Molality, m = 4.032 mol / 1.50 kg water = 2.688 mol kg^-1.
Freezing-point depression: ΔTf = Kf m = 1.86 × 2.688 = 5.00 K, so expected freezing point = -5.00 °C.
Boiling-point elevation: ΔTb = Kb m = 0.512 × 2.688 = 1.38 K, so expected boiling point = 101.38 °C.
The boiling specification is met, but the freezing specification is not: -5.0 °C is warmer than -8.0 °C.
For -8.0 °C, required m = 8.0 / 1.86 = 4.30 mol kg^-1. Required moles = 4.30 × 1.50 = 6.45 mol. Required mass = 6.45 × 62.0 = 400 g. Extra glycol needed ≈ 400 - 250 = 150 g. Real solutions may deviate from ideal behaviour, so the actual freezing point should be checked experimentally.
2. A student says molarity should be used because the density is given. Compare the molarity and molality values and explain which is appropriate for a colligative-property calculation here.
Read the worked answer
Total solution mass = 1500 g water + 250 g glycol = 1750 g. Volume = 1750 g / 1.02 g mL^-1 ≈ 1716 mL = 1.716 L. Molarity = 4.032 mol / 1.716 L = 2.35 mol L^-1.
Molality is 2.688 mol kg^-1. The two differ by about 14%, which is not negligible.
Colligative properties depend on the number of solute particles relative to the amount of solvent, and molality uses kilograms of solvent. It is also independent of temperature because it does not involve solution volume. Molarity would be convenient for volumetric work, but here it would underestimate the calculated freezing-point depression. Therefore molality is the appropriate concentration scale.
3. The workshop considers using NaCl instead of ethylene glycol. Would a 2.688 mol kg^-1 NaCl solution freeze below -8.0 °C? Explain using the van't Hoff factor, and state one practical reason the workshop might still avoid NaCl.
Read the worked answer
For NaCl, the ideal van't Hoff factor is i ≈ 2 because one formula unit gives Na+ and Cl-. Ideal ΔTf = i Kf m = 2 × 1.86 × 2.688 = 10.0 K, giving an ideal freezing point near -10.0 °C. This is below -8.0 °C, so on an ideal calculation NaCl would work at a lower molality than ethylene glycol.
However, in concentrated solution ion pairing makes the actual i less than 2, so the real freezing point would be warmer than -10.0 °C and should be measured. A practical objection is corrosion: chloride solutions are aggressive to many metals in a chiller, whereas ethylene glycol is commonly used with corrosion inhibitors. Thus NaCl is effective for freezing-point depression but may be unsuitable for the equipment.