1. Properties of Solutions Chapter 18 Lesson 3

2. Solution Composition Mass percentage (weight percentage): mass percentage of the component = X 100% mass of component total mass of mixture Mole fraction: The amount of a given component (in moles) divided by the total amount (in moles) X 1 = n 1 /(n 1 + n 2 ) for a two component system X 2 = n 2 /(n 1 + n 2 ) = 1 – X 1 or X 1 +X 2 =1 Mass Fraction, Mole Fraction, Molality and Molarity

3. Molality m solute = moles solute per kilogram solvent = moles per kg or (mol kg -1 ) Molarity (biochemists pay attention) c solute = moles solute per volume solution = moles per liter of solution (mol L -1 )

4. Colligative properties  Properties that depend upon the concentration of solute particles but not their identity Vapor pressure lowering Vapor pressure lowering Freezing point depression Freezing point depression Boiling point elevation Boiling point elevation Osmotic pressure Osmotic pressure

5. Raoult’s law  When nonvolatile solute is added to solvent, vapor pressure of solvent decreases in proportion to concentration of solute Freezing point goes down Freezing point goes down Boiling point goes up Boiling point goes up

6. Freezing and melting are dynamic processes  At equilibrium, rate of freezing = rate of melting

7. Units of concentration  Effect depends upon number of particles not mass of particles, so concentration must be in moles.  Molality (m) is used in these situations Moles solute/kg solvent Moles solute/kg solvent Temperature independent measure of concentration Temperature independent measure of concentration

8. Adding salts upsets the equilibrium  Fewer water molecules at surface: rate of freezing drops  Ice turns into liquid  Lower temperature to regain balance  Depression of freezing point

9. Freezing Point Depression: Solid/Liquid Equilibrium

10. When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = K f m solute (for nonelectrolytes) ΔT = freezing-point depression K f = freezing-point depression constant m solute = molality of solute Freezing-Point Depression

11. Example Question What is the freezing point of 1.40 mol Na 2 SO 4 in 1750 g H 2 O? Kf of H 2 O is 1.86 °C/m. -4.46 °C

12. Copyright © Houghton Mifflin Company. All rights reserved. 17a–12 NB: PROPERTY OF THE SOLVENT (NOT OF THE SOLUTE)

13. The same model explains elevated boiling point  Condensation and evaporation are dynamic processes  Replacing some of the liquid water with salt reduces rate of evaporation – leads to condensation  Raise temperature to recover balance

14. Boiling Point Elevation: Addition of a Solute

15. Nonvolatile solute elevates the boiling point of the solvent. ΔT = K b m solute ΔT = boiling-point elevation K b = boiling-point elevation constant m solute = molality of solute Boiling-Point Elevation

16. Changes in Boiling Point and Freezing Point of Water

17. Copyright © Houghton Mifflin Company. All rights reserved. 17a–17 Problem: A 0.0182-g sample of an unknown substance is dissolved in 2.135 g of benzene. The solution freezes at 5.14 o C instead of at 5.50 o C for pure benzene. K f (benzene) = 5.12 o C kg/mol What is the molecular weight of the unknown substance?

18. Copyright © Houghton Mifflin Company. All rights reserved. 17a–18 ΔT f = (5.50-5.14) = 0.36 o C ΔT f = K f m M = 0.36 o C / (5.12 o C kg/mol) = 0.070 m Solution: Freezing point depression  molality

19. Copyright © Houghton Mifflin Company. All rights reserved. 17a–19 Find the moles of unknown solute from the definition of molality: Mol solute = m x kg solvent = 0.070 mol 1 kg solvent x 0.002135 kg solvent = 1.5 x 10 -4 mol Solution:

20. Copyright © Houghton Mifflin Company. All rights reserved. 17a–20 Molar mass = 0.0182 g 1.5 x 10 -4 mol = 1.2 x10 2 g/mol (molecular weight = 1.2 x 10 2 g/mol) Solution: