1. Reading graph: at 38 °C the solubility of copper sulphate, CuSO 4, is 28g of anhydrous salt per 100g of water.

2. Reading graph: at 84 °C the solubility of potassium sulphate, K 2 SO 4, is 22g per 100g of water.

3. Ex Q1: How much potassium nitrate will dissolve in 20g of water at 34 °C? At 34 °C the solubility is 52g per 100g of water, so scaling down, 52 x 20 / 100 = 10.4g will dissolve in 20g of water.

4. Ex Q2: At 25 °C 6.9g of copper sulphate dissolved in 30g of water, what is its solubility in g/100cm 3 of water? Scaling up, 6.9 x 100 / 30 = 23g/100g of water (check on graph, just less than 23g/100g water).

5. http://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppt http://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppt Henry’s Law Application The solubility of pure N 2 (g) at 25 o C and 1.00 atm pressure is 6.8 x 10 -4 mol/L. What is the solubility of N 2 under atmospheric conditions if the partial pressure of N 2 is 0.78 atm? Step 1: Use the first set of data to find “k” for N 2 at 25°C Step 2: Use this constant to find the solubility (concentration) when P is 0.78 atm:

6. How do I get sugar to dissolve faster in my iced tea? Stir, and stir, and stir Add sugar to warm tea then add ice Grind the sugar to a powder Fresh solvent contact and interaction with solute Greater surface area, more solute-solvent interaction Faster rate of dissolution at higher temperature http://academic.pgcc.edu/~ssinex/Solutions.ppt

7. Revision Units of Concentrations amount of solute per amount of solvent or solution Percent (by mass) = g solute g solution x 100 g solute g solute + g solvent x 100 = Molarity (M) = moles of solute volume in liters of solution moles = M x V L http://academic.pgcc.edu/~ssinex/Solutions.ppt

8. Examples http://academic.pgcc.edu/~ssinex/Solutions.ppt What is the percent of KCl if 15 g KCl are placed in 75 g water? %KCl = 15g x 100/(15 g + 75 g) = 17% What is the molarity of the KCl if 90 mL of solution are formed? mole KCl = 15 g x (1 mole/74.5 g) = 0.20 mole molarity = 0.20 mole/0.090L = 2.2 M KCl

9. Gas Pressure and Solubility Quiz: The amount of dissolved oxygen in a mountain lake at 10,000 ft and 50 o F is __?_ than the amount of dissolved oxygen in a lake near sea level at 50 o F. Answer: Less at higher altitude because less pressure. A Coke at room temperature will have __?_ carbon dioxide in the gas space above the liquid than an ice cold bottle. Answer: More gas, because the warm coke can hold less of the gas in solution.

10. Gas Pressure and Solubility Hyperbaric therapy, which involves exposure to oxygen at higher than atmospheric pressure may be used to treat hypoxia (low oxygen supply in the tissues). Explain how the treatment works. Answer: The increase in pressure in the chamber will cause more gases to enter into lungs.

11. Calculate the Freezing Point of a 4.00 molal glycol/water solution. K f = 1.86 o C/molal (See K f table) Solution ∆T FP = K f m i = (1.86 o C/molal)(4.00 m)(1) = (1.86 o C/molal)(4.00 m)(1) ∆T FP = 7.44 FP = 0 – 7.44 = -7.44 o C (because water normally freezes at 0) Freezing Point Depression

12. At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆T FP = K f m i ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = 20.1 o C ∆T FP = 20.1 o C FP = 0 – 20.1 = -20.1 o C FP = 0 – 20.1 = -20.1 o C Freezing Point Depression

13. Calculate the Freezing Point of a 4.00 molal glycol/water solution. K f = 1.86 o C/molal (See K f table) Solution ∆T FP = K f m i = (1.86 o C/molal)(4.00 m)(1) = (1.86 o C/molal)(4.00 m)(1) ∆T FP = 7.44 FP = 0 – 7.44 = -7.44 o C (because water normally freezes at 0) Freezing Point Depression