Understanding Solutions and Colligative Properties: Key Concepts and Calculations

1. Chap13 - Solutions and Colligative Properties A solution is a solute (A) dissolved into a solvent (B).

2. Solute A Solvent B

3. A. Concentration • 1. Mass Percent = mass component x 100 total mass solution = grams A x 100 grams A + grams B

4. 2. Parts per million (ppm) = mass component x 106 total mass solution ppt = x 103 ( ) ppb = x 109 ( ) Parts per thousand Parts per billion

5. 3. Mole fraction (X) = mole component total moles XA = mole A mole A + mole B

6. 4. Molarity (M) = moles of solute Liters of solution M = mole A L solution *Remember* 1L = 1000mL Look at Problem #1 = 0.0200 M

7. 5. Molality (m) = moles solute kg solvent m = mole A kg B *Remember* 1g = 1mL for H2O 1000g = 1 kg Look at Problem #4 = 0.164 m

8. Dilution of Solution • M1V1 = M2V2 Answer to Worksheet 3a – 50 mL diluted 3b – 1.28 mL diluted

9. Conversion between Units • For H2O only, Molarity = molality. • Why? Because the density of H2O is equal to 1.00 g/mL. • Therefore, 1000mL = 1000g 1 L = 1 kg

10. Conversion between Units For any other solution other than an aqueous solution - YOU MUST USE THE DENSITY!!!!! Use the density to convert mass to volume.

11. Conversion between Units m = mole/kg mass solvent + add molar mass mass solute moles solute mass solution M = mole/L density volume solution

12. Conversion between Units m = mole/kg 1000 g = 1 kg mass solvent + add molar mass mass solute moles solute mass solution M = mole/L density volume solution 1000 mL = 1 L

13. Conversion between Units 1000 g = 1 kg + add molar mass density 1000 mL = 1 L

14. Your homework/classwork is worksheet –concentration conversions

15. Convert mass % to ….. • 5% HC2H3O2 5 g x 1mol/60g = 0.0833 mol • 95% H2O  95 g x 1mol/18g = 5.28 mol

16. Mole fraction • X = 0.0833mol / (0.0833 mol + 5.28 mol) • m = 0.0833 mol / 0.095 kg molality convert 95 g

17. Convert M to m • 1.13 mol to mol L solution kg solvent • 1000 mL x 1.05 g/ml = 1050 g solution • 1.13 mole KOH x 56.1 g/mol = 63.4 g solute density

18. 1050 g solution – 63.4 g KOH 986.6 g solvent • m = 1.13 mol KOH = 1.15 mol 0.9866 kg kg

19. What about mass percent? 63.4 g KOH 1050 g solution

20. Dimensional Analysis • What is the molarity of concentrated HCl? • 39.0% HCl by mass and 1.13 g/mL density

21. Solution Calculations • What is the molarity of a 1.11 ppm solution of Zn2+ ions?

22. Solid Calculations • Chemical analysis showed 1.23 mg Fe in a 15.67 g sample of soil. • What is the Fe concentration in ppm?

23. Unusual concentration units • How many nano moles of Cu are present in 12.3 µL of 25 ppm CuSO4?

24. B. Colligative Properties • 1. Boiling Point Elevation ΔTb = kb • m • i for an aqueous solution Tb = 100oC+ (0.52 oC/m) •(m) * note that as molality increases ΔTb increases as well Kbfor water Normal B.P.

25. B. Colligative Properties • 2. Freezing Point Depression ΔTf = kf • m • i for an aqueous solution Tf = 0oC- (1.86 oC/m) •(m) * note that as molality increases ΔTf increases as well Kf for water Normal F.P.

26. Ex. Non-electrolyte (i=1) • Antifreeze is made at 25% C2H6O2 by mass. What is the Tb and the Tf? • Make your life easy and assume 1000g. • Why? Because molality is based upon kg of solvent • Mass percent  250 g C2H6O2  750 g H2O

27. molality Boiling and Freezing Point C2H6O2 H2O

28. Ex. Molecular Weight of Unknown • What is the MM of a sample if 250grams of the sample is placed into 1000grams of water and the temperature rose by 3.5°C?

29. Assuming 1000g (1kg), the molality becomes…..

30. Ex. Electrolyte (i = ?) • IMPORTANT – the colligative properties of freezing point and boiling point are proportional to the number of particles present in the solution. • van Hoft factor, i • NaCl i = 2 moles Ie. 1m = 2m • CaCl2i = 3 moles Ie. 1m = 3m • Al2(SO4)3i = 5 moles Ie. 1m = 5m increasing colligative effect