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Understanding Colligative Properties: Impacts of Solute on Boiling and Freezing Points

In this activity, you'll partner up to explore a scenario where a solution's properties are affected by solute addition. You'll determine whether diluted HCl is suitable for laboratory use and calculate how solute particles influence boiling and freezing points. Learn to apply colligative properties by diving into examples such as NaCl and magnesium phosphate. This will enhance your understanding of how ionic versus covalent compounds behave when dissolved in solvents, along with key calculations for boiling point elevation and freezing point depression.

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Understanding Colligative Properties: Impacts of Solute on Boiling and Freezing Points

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  1. Dilution Review • Take 5 minutes with your partner to solve the scenario below. • Oops! I spilled 100mL of water into my 300mL of 6 M HCl. Is my solution too weak to use if my lab requires a 1M to 1.5 M HCl solution? • How do you know?

  2. Colligative Properties

  3. Defining Colligative Properties • PHYSICAL properties of the solvent that change when a solute is added • The degree of change is dependent upon the number of solute particles added to the solvent • The more solute particles = the greater the change in colligative property

  4. Number of Particles of Solute • Think back to what you’ve learned about ionic and covalent compounds dissolving in water. • Ionic compounds DISSOCIATE as they dissolve. • Covalent compounds do NOT dissociate as they dissolve. • Apply this idea. • How many solute particles will 1 piece of NaCl add to a solvent? • Answer: 2 (Na+ and Cl-)

  5. Applying the Concept Further • How many solute particles will 1 piece of magnesium phosphate add to a solvent? • Mg3(PO4)2 breaks into 3 Mg2+ ions and 2 PO43- ions. Answer: 5 solute particles • How many solute particles will 1 piece of sugar add to a solvent? • Sugar is covalent. Answer: 1 solute particle • Finally, which of the two above will affect a colligative property the most?

  6. Two Colligative Properties • Boiling point • Freezing point

  7. Boiling Point Elevation Boiling point is changed when solute is added to the solvent. • Boiling point elevates when a solute is added to a solvent. • The solution requires more energy to reach boiling. • Example: Salt water will not boil at 100°C. It will boil at a HIGHER temp.

  8. Calculating the NEW Boiling Pt. ∆Tb = Kbm i • ∆Tb = change in boiling pt. • Kb = boiling point constant for the solvent (will be given) • m = molality • i = number of ions present (USE ONLY WITH IONIC SOLUTES!)

  9. Example of Boiling Pt. Calculation • What is the boiling point when 15.0g NaCl is dissolved into 200 mL of water? (Kb of water is 0.52 °C/m) • ∆Tb = Kbm i • You are solving for ∆Tb , and you have the Kb to use. • m =You’ll need to calculate the molality from the info in the question. (change 15.0 g of NaCl to moles and 200 mL to kg and plug in) • i = Finally, is the solute ionic? YES…NaCl is ionic and will give TWO ions when dissolved.

  10. What is the boiling point when 15.0g NaCl is dissolved into 200 mL of water? • You’ll need to calculate molality first. • 15.0g NaCl x 1 mol NaCl = 0.256 moles NaCl 58.5 g NaCl • 200 mL = 200 g = 0.2 kg of water • Molality = 0.256 moles NaCl/ 0.2 kg water = 1.28m • ∆Tb = Kbm i • ∆Tb = (0.52 C/m)(1.28m)(2) • ∆Tb = 1.33 C (note: This is the CHANGE in boiling point. It does NOT answer our question.) • New Boiling Point: 100C + 1.33C = 101.33 • We took the regular boiling point of water and added the change since boiling point ELEVATES with a solute added.

  11. Ten Minute PartnersBoiling Point Calculations Kb of water = 0.52°C/m • What is the boiling point when 30.0g CaCl2 is dissolved into 200 mL of water? • The molality of the above example is similar to the problem that we worked together a moment ago. Explain the difference in boiling temperature. • CHALLENGING: How many grams of NaCl would need to be added to the water to change the boiling temperature of 200 mL to 110°C?

  12. Answers to Boiling Point Calculations 1. ∆Tb= (1.36m)(0.52C/m)(3) = 2.12C 100C + 2.12C = 102.12C 2. The difference is due to the difference in the number of solute particles (ions). • 10C = m(0.52)(2) m = 9.62 9.62m = mole/0.2kg mole = 1.92 mole NaCl 112.2 grams NaCl

  13. Freezing Point Depression Freezing point is changed when solute is added to the solvent. • Freezing point depresses when a solute is added to a solvent. • The solution requires a lower temp to reach freezing. • Example: Salt water will not freeze at 0°C. It will freeze at a LOWER temp.

  14. Calculating a NEW Freezing Pt. ∆Tf = Kfm i • ∆Tf = change in freezing pt. • Kf = freezing point constant for the solvent (will be given) • m = molality • i = number of ions present (USE ONLY WITH IONIC SOLUTES)

  15. Example of Freezing Pt. Calculation • What is the freezing point when 15.0g NaCl is dissolved into 200 mL of water? • (Kf of water is 1.86 °C/m) • ∆Tf = Kfm i • You are solving for ∆Tf , and you have the Kf to use. • m =You’ll need to calculate the molality from the info in the question. (change 15.0 g of NaCl to moles and 200 mL to kg and plug in) • i = Finally, is the solute ionic? YES…NaCl is ionic and will give TWO ions when dissolved.

  16. What is the freezing point when 15.0g NaCl is dissolved into 200 mL of water? • You’ll need to calculate molality first. • 15.0g NaCl x 1 mol NaCl = 0.256 moles NaCl 58.5 g NaCl • 200 mL = 200 g = 0.2 kg of water • Molality = 0.256 moles NaCl/ 0.2 kg water = 1.28m • ∆Tf = Kfm i • ∆Tf = (1.86 C/m)(1.28m)(2) • ∆Tf = 4.76 C (note: This is the CHANGE in freezing point. It does NOT answer our question.) • New Freezing Point: 0C – 4.76C = -4.76C • We took the regular freezing point of water and subtracted the change since freezing point DEPRESSES with a solute added.

  17. Ten Minute PartnersFreezing Point Calculations Kf of water = 1.86°C/m • What is the freezing point when 30.0g CaCl2 is dissolved into 200 mL of water? • The molality of the above example is similar to the problem that we solved together. Explain the difference in freezing temperature. • CHALLENGING: How many grams of NaCl would need to be added to the water to change the freezing temperature of 200 mL to -8°C?

  18. Answers to Freezing Point Calculations • ∆Tf= (0.271m)(1.86C/m)(3) =1.51C 0C - 1.51C = -1.51C 2. The difference is due to the difference in the number of solute particles (ions). • 8C = m(1.86C/m)(2) molality = 2.15 2.15 = mole/0.2kg mole = 0.43 mole NaCl 25.14 grams NaCl

  19. Practice Practice Practice! • The 15-4 Practice Problems in your practice packet are boiling point/freezing point calculations. • The key is posted at the front of the classroom. • PLEASE SOLVE MANY OF THESE PROBLEMS! You’ll see several on Wednesday’s test.

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