1. Expressing the Concentration of a Solute in a Solution

2. Several Ways of Expressing Concentration 1.mol/L  the most important way of expressing concentration in chemistry. Why? 2.% (m/v) 3.% (m/m)to be covered in 4.% (v/v) future lessons 5. ppm 6.ppb

3. 1.Molarity: mol solute per L solution units are: mol/L; mol·L -1 ; M or C = n/V, where n is # mol solute V is volume of solution (L) C is molarity of solute (mol/L) NOTE: Solution = solute + solvent

4. How to Prepare a Solution of Known Molarity That is, how would you prepare 250 mL of a solution of potassium nitrate, KNO 3, whose concentration (molarity) is 0.246 mol/L? C = n/V, C = 0.246 mol/L V = 0.250 L n = C*V = (0.246 mol·L -1 )(0.250 L) n = 0.0615 mol of KNO 3 is required. What now? Can we measure moles directly?

5. Need to convert mol to mass (g). mass = (0.0615 mol of KNO 3 )*(101.1 g/mol) mass KNO 3 = 6.22 g What must we do to get 6.22 g KNO 3 in 250 mL of solution? Do we simply add 6.22 g KNO 3 to 250 mL of water? No—solution is solute + solvent. How do we deal with this? We need special glassware...

6. Volumetric Flask A volumetric flask is specially designed to contain (TC) a certain volume of solution.

7. How to use a Volumetric Flask Put exact mass of solute in vol. flask; add some water to dissolve. Then fill to the line with water, inverting flask frequently to ensure good mixing. (Use a to add final few drops of water.)

8. Volumetric flasks come in many standard sizes:

9. Sample Problems 1. How would you prepare 100 mL of a solution that is 1.38 mol/L Na 2 SO 4. Include calculations and a description of how you would prepare the solution; mention any special glassware. C = 1.38 mol/L V = 0.100 L n = C*V = (1.38 mol/L)(0.100 L) = 0.138 mol... now convert this to grams

10. mass = (0.138 mol)*(142.1 g/mol) = 19.6 g Na 2 SO 4 Transfer 19.6 g Na 2 SO 4 to a 100 mL volumetric flask; dissolve; fill to the line with water. Use a medicine dropper to put the bottom of the meniscus on the line. the line 

11. 2. What is the molarity of a solution obtained by dissolving 6.58 g of glucose, C 6 H 12 O 6 in 50.0 mL of solution? C= n/V n= 6.58 g/(180.2 g/mol) = 0.0365 mol C= 0.0365 mol/0.0500 L = 0.730 mol/L (or M)

12. Now do the practice with a volumetric flask. Assignment due at end of class. HW. p 381 #41 to 50; do half a dozen (Quiz next class on this lesson. Includes volumetric flask practice and the HW.)