36 g of sucrose is mixed with 500 g of water. The ratio of freezing point depression to boiling point elevation will be [Kf of water = 1.86 K kg mol–1 and Kb of water = 0.52 K kg mol–1]
Question
36 g of sucrose is mixed with 500 g of water. The ratio of freezing point depression to boiling point elevation will be [Kf of water = 1.86 K kg mol–1 and Kb of water = 0.52 K kg mol–1]
Solution
First, we need to calculate the molality of the solution.
Molality (m) = moles of solute / mass of solvent in kg
The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol. So, the moles of sucrose in 36 g is 36 g / 342.3 g/mol = 0.105 mol.
The mass of water is 500 g, which is 0.5 kg.
So, the molality (m) = 0.105 mol / 0.5 kg = 0.21 mol/kg.
Next, we calculate the freezing point depression and boiling point elevation.
Freezing point depression = ΔTf = Kf * m Boiling point elevation = ΔTb = Kb * m
Where Kf is the molal freezing point depression constant for water (1.86 K kg mol–1) and Kb is the molal boiling point elevation constant for water (0.52 K kg mol–1).
So, ΔTf = 1.86 K kg mol–1 * 0.21 mol/kg = 0.39 K And, ΔTb = 0.52 K kg mol–1 * 0.21 mol/kg = 0.11 K
Finally, the ratio of freezing point depression to boiling point elevation is ΔTf / ΔTb = 0.39 K / 0.11 K = 3.55.
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