1. Chapter 8 Solutions
8.5 Molarity and Dilution

2. Molarity (M) Molarity (M) is a concentration term for solutions
gives the moles of solute in 1 L solution
molarity (M) = moles of solute
liter of solution

3. Preparing a 1.0 Molar NaCl Solution
A 1.0 M NaCl solution is prepared
by weighing out 58.5 g of NaCl (1.00 mole) and
adding water to make 1.0 liter of solution

4. Guide to Calculating Molarity

5. Example of Calculating Molarity
What is the molarity of L NaOH solution if it
contains 6.00 g of NaOH?
STEP 1 Given: 6.00 g of NaOH in L solution
Need: molarity (mole/L) of NaOH solution
STEP 2 Plan:
g of NaOH moles of NaOH molarity

6. Example of Calculating Molarity (continued)
STEP 3 Write equalitites and conversion factors:
1 mole of NaOH = g of NaOH
1 mole NaOH and g NaOH
40.0 g NaOH mole NaOH
STEP 4 Set up problem to calculate molarity:
6.00 g NaOH x 1 mole NaOH = mole of NaOH
40.0 g NaOH
0.150 mole NaOH = mole NaOH
0.500 L NaOH solution L NaOH solution
= M NaOH

7. Learning Check
What is the molarity of a solution if 325 mL of the solution contains 46.8 g of NaHCO3?
1) M NaHCO3
2) M NaHCO3
3) M NaHCO3

8. Solution 3) 1.71 M STEP 1 Given: 46.8 g of NaHCO3
325 mL (0.325 L) NaHCO3 solution
Need: molarity (mole/L) of NaHCO3 solution
STEP 2 Plan:
g of NaHCO moles of NaHCO molarity
STEP 3 Write equalities and conversion factors:
1 mole of NaHCO3 = g of NaHCO3
1 mole NaHCO3 and g NaHCO3
84.0 g NaHCO mole NaHCO3

9. Solution (continued)
STEP 4 Setup problem to calculate moles and molarity of NaHCO3:
46.8 g NaHCO3 x 1 mole NaHCO3
84.0 g NaHCO3
= mole of NaHCO3
0.557 mole NaHCO = mole NaHCO3
0.325 L NaHCO3 solution L NaHCO3 solution
= M NaHCO3

10. Learning Check
What is the molarity of a KNO3 solution if 225 mL of the solution contains 34.8 g of KNO3?
1) M
2) M
3) M

11. Solution
2) 1.53 M KNO3 STEP 1 Given: 34.8 g of KNO3 225 mL (0.225 L) KNO3 solution Need: molarity (mole/L) of KNO3 solution STEP 2 Plan: g of KNO3 moles of KNO3 molarity STEP 3 Write equalities and conversion factors: 1 mole of KNO3 = g of KNO3 1 mole KNO3 and g KNO g KNO3 1 mole KNO3

12. Solution (continued)
STEP 4 Set up problem to calculate moles and molarity of KNO3: 34.8 g KNO3 x 1 mole KNO g KNO3 = mole of KNO mole KNO3 = 1.53 mole KNO L KNO3 solution 1 L KNO3 solution = 1.53 M KNO3 In one setup: 34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M g KNO L

13. Molarity Conversion Factors
The units of molarity are used to write conversion factors for calculations with solutions.

14. Example of Calculations Using Molarity
How many grams of KCl are needed to prepare 125 mL
of a M KCl solution?
STEP 1 Given: 125 mL (0.125 L) of M KCl
Need: grams of KCl
STEP 2 Plan:
L of KCl moles of KCl g of KCl

15. Example of Calculations Using Molarity (continued)
STEP 3 Write equalities and conversion factors:
1 mole of KCl = g of KCl
1 mole KCl and g KCl
74.6 g KCl mole KCl
1 L of KCl = mole of KCl
1 L and mole KCl
0.720 mole KCl L
STEP 4 Set up problem to cancel mole KCl:
0.125 L x mole KCl x g KCl = 6.71 g of KCl
1 L mole KCl

16. Learning Check
How many grams of AlCl3 are needed to prepare 125 mL of a M solution?
1) g of AlCl3
2) g of AlCl3
3) g of AlCl3

17. Solution
3) 2.50 g of AlCl3 STEP 1 Given: 125 mL (0.125 L) of solution M AlCl3 solution Need: g of AlCl3 STEP 2 Plan: L of solution moles of AlCl3 g of AlCl3 STEP 3 Write equalities and conversion factors: 1 mole of AlCl3 = g of AlCl3 1 mole AlCl3 and g AlCl g AlCl3 1 mole AlCl3

18. Solution (continued) STEP 3 (continued) STEP 4 Set up problem:
1 L of KCl = mole of AlCl3
1 L and mole AlCl3
0.150 mole AlCl L
STEP 4 Set up problem:
0.125 L x mole AlCl3 x g = g of AlCl3
1 L mole AlCl3

19. Learning Check
How many milliliters of 2.00 M HNO3 contain 24.0 g of HNO3?
1) mL of 2.00 M HNO3
2) mL of 2.00 M HNO3
3) 190 mL of 2.00 M HNO3

20. Solution
3) 190 mL of HNO3 STEP 1 Given: 24.0 g of HNO M HNO3 solution Need: mL of HNO3 solution STEP 2 Plan: g of HNO3 moles of HNO3 L of HNO3 solution STEP 3 Write equalities and conversion factors: 1 mole of = 63.0 g of HNO3 1 mole HNO3 and 63.0 g HNO g HNO3 1 mole HNO3

21. Solution (continued)
STEP 4 Set up problem to calculate volume, in mL, of HNO3: 24.0 g HNO3 x 1 mole HNO3 x 1000 mL HNO g HNO moles HNO3 = 190 mL of a 2.00 M HNO3 solution

22. Dilution In a dilution, water is added volume increases
concentration of solute decreases

23. Initial and Diluted Solutions
In the initial and diluted solution,
the moles of solute are the same
the concentrations and volumes are related by the following equations:
For percent concentration
C1V = C2V2
Concentrated Diluted
solution solution
For molarity
M1V = M2V2

24. Guide to Calculating Dilution Quantities

25. Example of Dilution Calculations Using Percent Concentration
What volume of a 2.00% (m/v) HCl solution can be
prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?
STEP 1 Prepare a table:
C1 = 14.0% (m/v) V1 = mL
C2 = 2.00% (m/v) V2 = ?
STEP 2 Solve dilution expression for the unknown
C1V1 = C2V2 V2 = V1C1 C2
STEP 3 Set up the problem using known quantities:
V2 = V1C1 = (25.0 mL)(14.0%) = mL
C %

26. Learning Check What is the percent (m/v) of a solution prepared by
diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

27. Solution What is the percent (m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
STEP 1 Prepare a table:
C1 = % (m/v) V1 = mL
C2 = ? V2 = mL
STEP 2 Solve the dilution expression for the unknown: C1V1 = C2V2 C2 = C1V1 V2
STEP 3 Set up the problem using known quantities:
C2 = C1V1 = (10.0 mL)(9.00%) = 1.50% (m/v)
V mL

28. Example of Dilution Calculations Using Molarity
What is the molarity (M) of a solution prepared
by diluting L of M HNO3 to L?
STEP 1 Prepare a table:
M1 = M V1 = L
M2 = ? V2 = L
STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2
STEP 3 Set up the problem using known quantities:
M2 = M1V1 = (0.600 M)(0.180 L) = M
V L

29. Learning Check What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a M solution?
1) mL of a 1.80 M KOH
2) mL of a 1.80 M KOH
3) mL of a 1.80 M KOH

30. Solution What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a M solution?
STEP 1 Prepare a table:
M1 = M V1 = mL
M2 = M V2 = ?
STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2
STEP 3 Set up the problem using known quantities:
V2 = M1V1 = (1.80 M)(15.0 mL) = 90.0 mL
M M

31. Molarity in Chemical Reactions
In a chemical reaction,
the volume and molarity of a solution are used to determine the moles of a reactant or product
molarity ( mole ) x volume (L) = moles
1 L
if molarity (mole/L) and moles are given, the volume (L) can be determined
moles x L = volume (L)
moles

32. Guide to Calculations Involving Solutions in Chemical Reactions

33. Using Molarity of Reactants
How many mL of 3.00 M HCl are needed to completely
react with 4.85 g of CaCO3?
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
STEP 1 Given: M HCl; 4.85 g of CaCO3
Need: volume of HCl in mL
STEP 2 Write a plan:
g of CaCO moles of CaCO moles of HCl mL of HCl
STEP 3 Write equalities and conversion factors:
1 mole of CaCO3 = g
1 mole CaCO and g CaCO3
100.1 g CaCO mole CaCO3

34. Using Molarity of Reactants (continued)
STEP 3 (continued)
1 mole of CaCO3 = 2 moles of HCl
1 mole of CaCO3 and 2 moles of HCl
2 moles of HCl mole of CaCO3
1000 mL of HCl = 3.00 moles of HCl
1000 mL of HCl and moles of HCl
3.00 moles of HCl mL of HCl
STEP 4 Set up problem to calculate mL of HCl:
4.85 g CaCO3 x 1 mole CaCO3 x 2 moles HCl x 1000 mL HCl
100.1 g CaCO mole CaCO moles HCl
= mL of HCl required

35. Learning Check
If 22.8 mL of M MgCl2 is needed to completely react 15.0 mL of AgNO3 solution, what is the molarity of the AgNO3 solution?
MgCl2(aq) + 2AgNO3(aq) AgCl(s) + Mg(NO3)2(aq)
1) M
2) M
3) M

36. Solution
3) M AgNO3
STEP 1 Given: mL (0.228 L) of M MgCl Need: molarity of AgNO3
STEP 2 Write a plan to calculate molarity:
mL of MgCl moles of MgCl moles of AgNO molarity of AgNO3
STEP 3 Write equalities and conversion factors:
0.100 mole of MgCl2 = 1 L of MgCl2
0.100 mole MgCl2 and 1 L MgCl L MgCl mole MgCl2

37. Solution (continued) STEP 3 (continued)
1 mole of MgCl2 = 2 moles AgNO3
2 moles AgNO and 1 mole MgCl2
1 mole MgCl moles AgNO3
STEP 4 Set up problem to calculate molarity of AgNO3:
L x mole MgCl2 x 2 moles AgNO3 x ____
1 L mole MgCl L
= mole/L = M AgNO3

38. Learning Check
How many liters of H2 gas at STP are produced when 125 mL of 6.00 M HCl reacts with sufficient Zn?
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
1) L of H2
2) L of H2
3) L of H2

39. Solution
2) L of H2 gas
STEP 1 Given: 125 mL (0.125 L) of 6.00 M HCl Need: liters of H2 at STP
STEP 2 Write a plan to calculate liters of H2:
L of HCl moles of HCl moles of H liters of H2
STEP 3 Write equalities and conversion factors:
6.00 moles of HCl = 1 L of HCl
6.00 moles HCl and L HCl
1 L HCl moles HCl
22.4 L of H2 = 1 mole of H2 at STP
22.4 L H and mole H2
1 mole H L H2

40. Solution (continued) STEP 3 (continued) 2 moles HCl = 1 mole H2
2 moles HCl and mole H2
1 mole H moles HCl
STEP 4 Set up problem to calculate liters of H2:
0.125 L x 6.00 moles HCl x 1 mole H2 x L
1 L moles HCl 1 mole H2
= L of H2 gas