1. Solutions
Molarity

2. Molarity (M) A concentration that expresses the
moles of solute in 1 L of solution
Molarity (M) = moles of solute
1 liter solution

3. Molarity Conversion Factors
A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors.
3.0 moles NaOH and L NaOH soln
1 L NaOH soln moles NaOH
LecturePLUS Timberlake

4. Units of Molarity 2.0 M HCl = 2.0 moles HCl 1 L HCl solution

5. Molarity Calculation
If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

6. Calculating Molarity = 0.20 M NaOH
1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH
40.0 g NaOH
2) 500. mL x 1 L _ = L
1000 mL
mole NaOH = mole NaOH
0.500 L L
= 0.20 M NaOH

7. Check for Understanding
1) A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution?
1) 8 M
2) 5 M
3) 2 M
Drano

8. Solution #1
A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution?
2) 5 M
M = 2 mole KOH = 5 M
0.4 L
Drano

9. Checking for Understanding
2) A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution?
1) 0.20 M
2) 5.0 M
3) 36 M

10. Solution #2
A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution?
1) 72 g x 1 mole x = M
180. g L

11. M1V1 = M2V2 Making dilutions Equation explained….
M1 x V1 = moles before dilution
Moles before dilution = moles after dilution
M2 x V2 = moles after dilution

12. Sample problem 4.7 (from textbook)
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? Known: M1 = 16 M M2 = 0.10 M V2 = 1.5 L Unknown: V1 M1V1 = M2V2  16M x V1 = 0.10 M x 1.5 L 0.10 M x 1.5 L = 0.15 mol V1 = 0.15 mol 16 M = L or 9.4 mL H2SO4

13. Practice!
What volume of 12.0 M HCl would you need to prepare 200 mL of 3.0 M HCl?