Q3. What is the freezing point depression of a solution containing 3 moles of KCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol) (2022)
Solution:
Freezing point depression = i * Kf * m = 3 * 1.86 * 3 = 5.58 °C (i = 2 for KCl).
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Q4. What is the vapor pressure of a solution containing 2 moles of solute and 4 moles of solvent compared to pure solvent? (2022)
Solution:
Vapor pressure lowering = (moles of solute / total moles) = 2 / (2 + 4) = 2/6 = 1/3. Thus, vapor pressure is 1/2 of pure solvent.
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Q5. What is the vapor pressure of a solution containing 1 mole of solute in 9 moles of solvent, assuming ideal behavior? (2019)
Solution:
Using Raoult's law, the vapor pressure of the solution = (moles of solvent / total moles) * P0 = (9 / 10) * P0 = 0.9 P0.
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Q6. If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.52 °C kg/mol) (2021)
Solution:
Boiling point elevation = i * Kb * m = 1 * 0.52 * 0.5 = 0.26 °C, so change = 0.52 °C.
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Q7. If 1 mole of a non-volatile solute is dissolved in 1 kg of water, what is the expected freezing point depression? (2019)
Solution:
Freezing point depression (ΔTf) = i * Kf * m = 1 * 1.86 °C kg/mol * 1 mol/kg = 1.86 °C.
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Q8. What is the boiling point elevation for a solution with 0.5 moles of a non-volatile solute in 1 kg of water? (Kb for water = 0.512 °C kg/mol) (2022)
Solution:
Boiling point elevation (ΔTb) = Kb * m = 0.512 °C kg/mol * 0.5 mol/kg = 0.256 °C.
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Q9. If 1 kg of water is mixed with 1 kg of sugar, what is the expected change in boiling point? (2023)
Solution:
The addition of sugar (a non-volatile solute) will raise the boiling point of water due to boiling point elevation.
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Q10. What is the osmotic pressure of a solution containing 0.1 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol)) (2020)
