Sample problems colligative properties of solutions

9 questions about Sample problems colligative properties of solutions

  1. Question: What is the boiling point of a solution prepared by dissolving 20 g of glucose (molar mass = 180 g/mol) in 500 g of water (Kb = 0.512°C/m)? Solution: molality = moles solute/kg solvent = (20g / 180 g/mol) / (500g / 1000g/kg) = 0.222 m; ΔTb = Kb x molality = 0.512°C/m x 0.222 m = 0.114°C; The boiling point of the solution is 100°C + 0.114°C = 100.114°C.

  2. Question: Calculate the freezing point of a solution containing 5 g of NaCl (molar mass = 58.44 g/mol) in 200 g of water (Kf = 1.86°C/m). Solution: Since NaCl dissolves as two particles (Na⁺ and Cl⁻), the van ‘t Hoff factor (i) is 2. Molality = moles solute/kg solvent = (5g / 58.44 g/mol) / (200g / 1000g/kg) = 0.429 m; ΔTf = i x Kf x molality = 2 x 1.86°C/m x 0.429 m = 1.59°C; The freezing point of the solution is 0°C – 1.59°C = -1.59°C.
  3. Question: How many grams of ethylene glycol (C₂H₆O₂, molar mass = 62.07 g/mol) must be added to 500 g of water to raise the boiling point by 1.5°C (Kb = 0.512°C/m)? Solution: ΔTb = Kb x molality; 1.5°C = 0.512°C/m x molality; molality = 2.93 m. Moles solute = molality x kg solvent = 2.93 mol/kg x 0.500 kg = 1.465 mol. Mass solute = moles solute x molar mass = 1.465 mol x 62.07 g/mol = 91 g.
  4. Question: Calculate the osmotic pressure at 25°C of a solution prepared by dissolving 1.5 g of CaCl₂ (molar mass = 110.98 g/mol) in 250 mL of water. Solution: Moles of solute = 1.5g / 110.98 g/mol = 0.014 mol. Since CaCl₂ dissolves as three particles (Ca²⁺ and 2Cl⁻), the van ‘t Hoff factor (i) is 3. Molarity = moles solute/L solution = 0.014 mol / 0.250 L = 0.054 M. Osmotic pressure = i x M x R x T = 3 x 0.054 M x 0.0821 L.atm/K.mol x (25°C + 273) K = 3.6 atm.
  5. Question: What is the vapor pressure at 25°C of a solution prepared by dissolving 10 g of glucose (C₆H₁₂O₆, molar mass = 180 g/mol) in 90 g of water, if the vapor pressure of pure water at 25°C is 23.8 mmHg? Solution: Moles of solute = 10g / 180 g/mol = 0.056 mol. Moles of water = 90g / 18.02 g/mol = 5 mol. Mole fraction of water = moles water / (moles water + moles solute) = 5 / (5 + 0.056) = 0.989. Vapor pressure of solution = mole fraction water x vapor pressure water = 0.989 x 23.8 mmHg = 23.5 mmHg.
  6. Question: What is the freezing point of a solution prepared by dissolving 15 g of MgCl₂ (molar mass = 95.21 g/mol) in 200 g of water (Kf = 1.86°C/m)? Solution: Since MgCl₂ dissolves as three particles (Mg²⁺ and 2Cl⁻), the van ‘t Hoff factor (i) is 3. Molality = moles solute/kg solvent = (15g / 95.21 g/mol) / (200g / 1000g/kg) = 0.079 m; ΔTf = i x Kf x molality = 3 x 1.86°C/m x 0.079 m = 0.44°C; The freezing point of the solution is 0°C – 0.44°C = -0.44°C.
  7. Question: What is the boiling point of a solution prepared by dissolving 30 g of urea (CO(NH₂)₂, molar mass = 60 g/mol) in 500 g of water (Kb = 0.512°C/m)? Solution: molality = moles solute/kg solvent = (30g / 60 g/mol) / (500g / 1000g/kg) = 1 m; ΔTb = Kb x molality = 0.512°C/m x 1 m = 0.512°C; The boiling point of the solution is 100°C + 0.512°C = 100.512°C.
  8. Question: Calculate the freezing point of a solution containing 8 g of KBr (molar mass = 119.01 g/mol) in 300 g of water (Kf = 1.86°C/m). Solution: Since KBr dissolves as two particles (K⁺ and Br⁻), the van ‘t Hoff factor (i) is 2. Molality = moles solute/kg solvent = (8g / 119.01 g/mol) / (300g / 1000g/kg) = 0.022 m; ΔTf = i x Kf x molality = 2 x 1.86°C/m x 0.022 m = 0.082°C; The freezing point of the solution is 0°C – 0.082°C = -0.082°C.
  9. Question: How many grams of sodium acetate (CH₃COONa, molar mass = 82.03 g/mol) must be added to 500 g of water to lower the freezing point by 1.5°C (Kf = 1.86°C/m)? Solution: ΔTf = Kf x molality; 1.5°C = 1.86°C/m x molality; molality = 0.806 m. Moles solute = molality x kg solvent = 0.806 mol/kg x 0.500 kg = 0.403 mol. Mass solute = moles solute x molar mass = 0.403 mol x 82.03 g/mol = 33 g.

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