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

3. Solvent B Solute A

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

5. 2. Parts per million (ppm) 2. Parts per million (ppm) = mass componentx 10 6 total mass solution total mass solution ppt = x 10 3 ( ) ppb = x 10 9 ( ) Parts per thousand Parts per billion

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

7. 4. Molarity (M) 4. Molarity (M) = moles of solute Liters of solution Liters of solution M = mole A L solution L solution 3.16 g MgBr 2 1 mol= 0.0172 mol MgBr 2 184.1 g MgBr 2 *Remember* 1L = 1000mL 0.859 L Look at Problem #1 = 0.0200 M

8. 5. Molality (m) 5. Molality (m) = moles solute kg solvent kg solvent m = mole A kg B kg B *Remember* 1g = 1mL for H 2 O 1000g = 1 kg 4.8 g NaCL1 mol= 0.0821 mol NaCl 58.5 g NaCl 0.5 kg Look at Problem #4 = 0.164 m

9. Dilution of Solution M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2 Answer to Worksheet 3a – 50 mL diluted 3b – 1.28 mL diluted

10. Conversion between Units For H 2 O only, Molarity = molality. For H 2 O only, Molarity = molality. Why? Because the density of H 2 O is equal to 1.00 g/mL. Why? Because the density of H 2 O is equal to 1.00 g/mL. Therefore, 1000mL = 1000g Therefore, 1000mL = 1000g 1 L = 1 kg 1 L = 1 kg

11. 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.

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

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

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

15. Your homework/classwork is worksheet – concentration conversions

16. Convert mass % to ….. Convert mass % to ….. 5% HC 2 H 3 O 2  5 g x 1mol/60g = 5% HC 2 H 3 O 2  5 g x 1mol/60g = 0.0833 mol 95% H 2 O  95 g x 1mol/18g = 95% H 2 O  95 g x 1mol/18g = 5.28 mol

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

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

19. 1050 g solution 1050 g solution – 63.4 g KOH – 63.4 g KOH 986.6 g solvent 986.6 g solvent m = 1.13 mol KOH = 1.15 mol m = 1.13 mol KOH = 1.15 mol 0.9866 kg kg 0.9866 kg kg

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

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

22. Solution Calculations What is the molarity of a 1.11 ppm solution of Zn 2+ ions? What is the molarity of a 1.11 ppm solution of Zn 2+ ions?

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

24. Unusual concentration units How many nano moles of Cu are present in 12.3 µL of 25 ppm CuSO 4 ? How many nano moles of Cu are present in 12.3 µL of 25 ppm CuSO 4 ?

25. B. Colligative Properties 1. Boiling Point Elevation 1. Boiling Point Elevation ΔT b = k b m i for an aqueous solution T b = 100 o C + (0.52 o C/m) (m) * note that as molality increases * note that as molality increases ΔT b increases as well ΔT b increases as well Normal B.P. K b for water

26. B. Colligative Properties 2. Freezing Point Depression 2. Freezing Point Depression ΔT f = k f m i for an aqueous solution T f = 0 o C - (1.86 o C/m) (m) * note that as molality increases * note that as molality increases ΔT f increases as well ΔT f increases as well Normal F.P. K f for water

27. Ex. Non-electrolyte (i=1) Antifreeze is made at 25% C 2 H 6 O 2 by mass. What is the T b and the T f ? Antifreeze is made at 25% C 2 H 6 O 2 by mass. What is the T b and the T f ? Make your life easy and assume 1000g. Make your life easy and assume 1000g. Why? Because molality is based upon kg of solvent Why? Because molality is based upon kg of solvent Mass percent  250 g C 2 H 6 O 2 Mass percent  250 g C 2 H 6 O 2  750 g H 2 O

28. Boiling and Freezing Point molality C2H6O2C2H6O2 H2OH2O

29. 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? 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?

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

31. Ex. Electrolyte (i = ?) IMPORTANT – the colligative properties of freezing point and boiling point are proportional to the number of particles present in the solution. 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 van Hoft factor, i NaCl  i = 2 moles Ie. 1m = 2m NaCl  i = 2 moles Ie. 1m = 2m CaCl 2  i = 3 moles Ie. 1m = 3m CaCl 2  i = 3 moles Ie. 1m = 3m Al 2 (SO 4 ) 3  i = 5 moles Ie. 1m = 5m Al 2 (SO 4 ) 3  i = 5 moles Ie. 1m = 5m increasing colligative effect