1. Chapter 13 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. Units of Molarity 2.0 M HCl = 2.0 moles HCl 1 L HCl solution
6.0 M HCl = moles L HCl solution

4. Math Problems Molarity is a conversion factor 3 types of M problems
if you know the molarity
1 - You can determine the # of moles (or grams) solute in a given volume of solution.
2 - You can calculate the vol of solution that contains a specific # of moles solute if you know the molarity
3 If given the moles (grams) solute and vol of solution
determine the molarity

5. Molarity Calculation NaOH is used to open stopped sinks, to treat
cellulose in the making of nylon, and to
remove potato peels commercially.
If 4.0 g NaOH (mm = 40.0 g/mol) are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?
1 - calc the moles solute
Mol = g/mm
2 - Determine L of solution
L = mL/1000
3 - M = mol solute/L solution

6. Calculating Molarity 1. 4.0 g NaOH x 1 mol NaOH = 0.10 mol NaOH
mL x 1 L = L
1000 mL
mol NaOH = mol NaOH
0.500 L L
= M NaOH

7. Learning Check M1
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

8. Learning Check M1
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

9. Learning Check
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 - Determine moles of solute (glucose)
2 - Divide by L solution
1) 0.20 M
2) 5.0 M
3) 36 M

10. Solution
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 = mol = M
180. g L

11. Molarity Conversion Factors
A solution is a 3.0 M NaOH. Write the molarity in the form of conversion factors that relates vol of solution to # of mol solute
3.0 moles NaOH and 1 L NaOH soln
1 L NaOH soln moles NaOH

12. Learning Check
Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution?
1 - determine L of solution
2 - convert from L to moles with molarity
1) 15 moles HCl
2) moles HCl
3) moles HCl

13. Solution
3) mL x L = L
1000 mL
1.5 L x mole HCl = 0.15 mole HCl
1 L
(Molarity is a conversion factor)

14. Learning Check
How many grams of KCl are present in 2.5 L of 0.50 M KCl?
1 - convert from L to moles solute, KCl
2 - convert moles to grams of KCl
1) 1.3 g
2) 5.0 g
3) 93 g

15. Solution 3)
2.5 L x mole x g KCl = 93 g KCl
1 L mole KCl

16. Learning Check M5
How many liters of stomach acid, which is 0.10 M HCl, contain 0.15 mole HCl?
1 - Convert from mol to L with molarity
1) L
2) L
3) L

17. Solution M5 0.15 mol HCl x 1 L soln 0.10 mol HCl (Molarity inverted)2)
0.15 mol HCl x 1 L soln
0.10 mol HCl
(Molarity inverted)
= L HCl

18. Learning Check M6
How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?
1 - Deter L of solution
2 - Convert L to moles with molarity
3 - Convert mol to grams with molar mass
1) 12 g
2) 48 g
3) 300 g

19. Solution
2) mL x 1 L = L
1000 mL
0.400 L x 3.0 mol NaOH x g NaOH
1 L mol NaOH
(molarity) (molar mass)
= 48 g NaOH