Answer :
Final answer:
The freezing point of a 16.1 g NaCl in 139 mL water solution would be -2.46 °C.
Explanation:
The freezing point of a solution is lowered when a solute, such as NaCl, is added to a solvent, such as water. To calculate the freezing point depression, we can use the formula ΔT = Kf × m, where ΔT is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solution.
First, we need to calculate the molality of the solution. Molality (m) is the moles of solute divided by the mass of the solvent in kilograms. The molar mass of NaCl is 58.44 g/mol.
m = (16.1 g NaCl / 58.44 g/mol) / (0.139 kg H2O) = 1.32 mol/kg
Next, we can use the cryoscopic constant (Kf) for water, which is 1.86 °C/m, to calculate the freezing point depression:
ΔT = (1.86 °C/m) × (1.32 mol/kg) = 2.46 °C
The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution would be 0 °C - 2.46 °C = -2.46 °C.
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