1. Solutions Chapter 16

2. Solutions A solution is a homogenous mixture of 2 substances.

3. Solvent and Solute The solvent is the dissolving medium in a solution. The solute are the particles that are dissolved in the solution.

4. Rate of dissolution 3 factors affect how fast something dissolves: – Stirring – Temperature – Surface area (size) of the solute particles

5. Solubility The solubility of a substance is the amount of solute that dissolves in a given quantity of solvent at a specific temperature and pressure.

6. Saturation A saturated solution contains the maximum amount of solute possible. An unsaturated solution is a solution that contains less solute than can possibly be dissolved. A supersaturated solution contains more solute than it can theoretically hold.

7. Can they dissolve? If 2 liquids can dissolve in each other, they are miscible (water and vinegar) If 2 liquids cannot dissolve in each other, they are immiscible (oil and water)

8. Molarity Molarity (M) is the number of moles of solute dissolved in one liter (1 L) of solution. To calculate the molarity of a solution, divide the moles of solute by the volume of solution: Molarity (M) = moles of solute liters of solution

9. Practice An IV bag has 0.90 g NaCl in 0.1 L of solution. What is its molarity? 1. Convert grams to moles. 2. Calculate molarity.

10. Practice An IV bag has 0.90 g NaCl in 0.1 L of solution. What is its molarity? 1. Convert grams to moles. 0.90 g NaCl x 1 mol NaCl= 1 58.443 g NaCl

11. Practice An IV bag has 0.90 g NaCl in 0.1 L of solution. What is its molarity? 1. Convert grams to moles. 0.90 g NaCl x 1 mol NaCl= 0.01539962 mol 1 58.443 g NaCl Don’t round until the end!

12. Practice An IV bag has 0.90 g NaCl in 0.1 L of solution. What is its molarity? 2. Calculate molarity. (moles/liters)

13. Practice An IV bag has 0.90 g NaCl in 0.1 L of solution. What is its molarity? 2. Calculate molarity. (moles/liters) 0.01539962 mol NaCl = 0.1L

14. Practice An IV bag has 0.90 g NaCl in 0.1 L of solution. What is its molarity? 2. Calculate molarity. (moles/liters) 0.01539962 mol NaCl = 0.153996201 = 0.2 M 0.1L

15. Practice Again A solution has a volume of.25L and contains 0.70 mol of salt. What is its molarity (M)? Molarity = moles liters

16. Practice Again A solution has a volume of.25L and contains 0.70 mol of salt. What is its molarity (M)? Molarity = 0.70 mol 0.25 liters=

17. Practice Again A solution has a volume of.25L and contains 0.70 mol of salt. What is its molarity (M)? Molarity = 0.70 mol 0.25 liters= 2.8 M

18. Solubility Lab Today: Complete your – Title – Intro (2 paragraphs) – Hypothesis – Materials – Methods – Data Table Next class: conduct the lab, then complete – Data Table – Graph – Conclusion (3 paragraphs) – Literature Cited

19. Colligative Properties of Solutions A property that depends only upon the number of solute particles, and not upon their identity, is called a colligative property.

20. 3 Colligative Properties 1. Vapor Pressure Lowering – Vapor pressure is the pressure exerted by a vapor on a liquid in a closed system – A solution always has a lower vapor pressure than a pure solvent – Example: salt water has a lower vapor pressure than pure water

21. 3 Colligative Properties 2. Boiling-point elevation – The boiling point of a substance is the temperature at which it turns from liquid to gas – A solution always has a higher boiling point than a pure solvent – Example: salt water has a higher boiling point than pure water

22. 3 Colligative Properties 3. Freezing Point Depression – The freezing point of a substance is the temperature when the liquid becomes solid. – A solution always has a lower freezing point than a pure solvent – Example: salt water has a lower freezing point than pure water

23. Molality (m) The unit molality (m) is the number of moles of solute dissolved in one kilogram (1kg) of solvent. It is also called molal concentration. Molality = moles of solute kilograms of solvent

24. Practice Calculate the molality of a solution prepared by dissolving 10.0 moles of salt in 0.60 kg of water. Molality = moles of solute kilograms of solvent

25. Practice Calculate the molality of a solution prepared by dissolving 10.0 moles of salt in 0.60 kg of water. Molality = 10.0 moles NaCl = 0.60 kg of water

26. Practice Calculate the molality of a solution prepared by dissolving 10.0 moles of salt in 0.60 kg of water. Molality = 10.0 moles NaCl = 16.6666667 0.60 kg of water = 17 m

27. Practice again How many moles of potassium iodide (KI) must be dissolved in 0.50 kg of water to produce a 0.060 molal KI solution? Molality = moles of solute kilograms of solvent

28. Practice again How many moles of potassium iodide (KI) must be dissolved in 0.50 kg of water to produce a 0.060 molal KI solution? 0.060 m = moles of solute = 0.50 kg water

29. Practice again How many moles of potassium iodide (KI) must be dissolved in 0.50 kg of water to produce a 0.060 molal KI solution? 0.060 m = moles of solute = 0.03 moles KI 0.50 kg water

30. Last practice Calculate the molality of a solution prepared by dissolving 20.0 g of salt in 600 g of water. 1000 g = 1 kg 1. Convert grams of salt to moles of salt 2. Convert grams of water to kilograms of water 3. Calculate molality

31. Last practice Calculate the molality of a solution prepared by dissolving 20.0 g of salt in 600 g of water. 1000 g = 1 kg 1. Convert grams of salt to moles of salt 20.0 g NaCl x 1 mol NaCl= 0.342213781 mol NaCl 1 58.443 g NaCl

32. Last practice Calculate the molality of a solution prepared by dissolving 20.0 g of salt in 600 g of water. 1000 g = 1 kg 2. Convert grams of water to kilograms of water 600 g water x 1 kg water = 1 1000 g water

33. Last practice Calculate the molality of a solution prepared by dissolving 20.0 g of salt in 600 g of water. 1000 g = 1 kg 2. Convert grams of water to kilograms of water 600 g water x 1 kg water = 0.6 kg water 1 1000 g water

34. Last practice Calculate the molality of a solution prepared by dissolving 20.0 g of salt in 600 g of water. 1000 g = 1 kg 3. Calculate molality Molality = moles of solute kilograms of solvent

35. Last practice Calculate the molality of a solution prepared by dissolving 20.0 g of salt in 600 g of water. 1000 g = 1 kg 3. Calculate molality Molality = 0.342213781 mol NaCl = 0.6 kg water

36. Last practice Calculate the molality of a solution prepared by dissolving 20.0 g of salt in 600 g of water. 1000 g = 1 kg 3. Calculate molality Molality = 0.342214781 mol NaCl = 0.570356302 m 0.6 kg water = 0.6 m