What would the final freezing point of water be if 3 mol of sugar were added to 1 kg of water (Kf = 1.86C/(mol/kg) for water and i = 1 for sugar)?A.-0.62CB.-1.86CC.+5.58CD.-5.58C
Question
What would the final freezing point of water be if 3 mol of sugar were added to 1 kg of water (Kf = 1.86C/(mol/kg) for water and i = 1 for sugar)?A.-0.62CB.-1.86CC.+5.58CD.-5.58C
Solution
The freezing point depression, ΔTf, is calculated using the formula:
ΔTf = i * Kf * m
where: i = van 't Hoff factor (the number of particles the solute splits into, which is 1 for sugar) Kf = cryoscopic constant (which is 1.86°C/m for water) m = molality (moles of solute per kilogram of solvent)
Given that 3 mol of sugar are added to 1 kg of water, the molality (m) is 3 mol/kg.
Substituting the given values into the formula, we get:
ΔTf = 1 * 1.86°C/m * 3 m = 5.58°C
The freezing point of pure water is 0°C. The freezing point of the solution will be lower by 5.58°C due to the added sugar. Therefore, the final freezing point of the water will be:
0°C - 5.58°C = -5.58°C
So, the correct answer is D. -5.58°C.
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