Final answer:
The freezing point of 4.38 kg of pure water is 0.0°C, which is consistent for any quantity of pure water at standard atmospheric pressure.
Step-by-step explanation:
The freezing point of pure water is 0.0°C under standard atmospheric pressure, which is the same for any amount of water, regardless of the mass, as long as it is pure and the pressure is one atmosphere. However, if we are considering a situation where the freezing point of a solution is reduced to 24.0°C due to the presence of a solute, then we must be talking about a colligative property known as freezing point depression. In such a case, the presence of a solute would cause the freezing point to be lowered, which is calculated using the formula ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van't Hoff factor (ionization factor), Kf is the cryoscopic constant, and m is the molality of the solution.
Since the given question does not mention the presence of a solute, we can assume that the water is pure. Therefore, the freezing point remains the standard 0°C for the 4.38 kg of water the student asked about, given that there are no impurities or pressure changes.