This document provides 11 practice problems for calculating molarity, molality, and normality. It gives the steps to take for each problem, including identifying key information, setting up the appropriate equation, showing the calculations, and including the correct units in the answer. The problems cover a range of scenarios involving determining molarity or molality based on moles of solute and volume of solution or solvent.

1. Practice Problems, Part 1: Molarity, molality, Normality
Solve each problem on your own paper. Be sure to show:
• Information the problem gives you- numbers and its description
• Set up of the math
• Doing the math; use dimensional analysis whenever possible
• Answer before rounding
• Answer after rounding
• No naked numbers- put units on all numbers
1. Calculate the Molarity of a solution that has 2.3 moles dissolved in 4.6
liters total.
2.3 moles
4.6 liters of solution
Need to solve for Molarity
Moles =?
Liters
2.3 moles = 0.50 moles/L = 0.50M
4.6 liters
2. Calculate the molality of a solution that has 1.5 moles added to 675 mL of
solvent.
1.5 moles
675 mL of solvent
Need to convert mL to L
Need to solve moles/L
675 mL x 1L = 0.675 L
1000mL
1.5 moles = 2.2222 moles =2.2 moles/liter = 2.2 m
0.675 L 1 L of solvent
3. Calculate the Molarity of 23.5 grams of silver nitrate, AgNO 3, that is
dissolved in water for a final volume of 750 mL.
23.5 g AgNO3
750 mL final volume
Need to convert g to moles
Need to convert 750 mL to L
Getz 2007

2. Need to use equation moles/liter
Ag = 107.9 g/mol
N = 14.0 g/mol
3 oxygens = 3 (16.0) = 48.0 g/mol
107.9 + 14.0 + 48.0 = 169.9 g/mol
23.5 grams x 1 mol = 0.1383 mol = 0.138 mol
169.9 g
750 mL x 1L = 0.75 L
1000 mL
0.138 mol = 0.1844 mol/L = 0.18 mol/L = 0.18 M
0.75 L
4. How many grams of CuSO4 ⋅ 5H2O do you need to make 800. mL of a 0.6 M
solution? Remember to calculate the molar mass of this chemical you figure
out the mass of Cu, S, four oxygens, and five waters. You treat the ⋅ as a +
not as an x.
800. mL
0.6 M
Need grams of CuSO4 ⋅ 5H2O
Need to figure out mol of CuSO4 ⋅ 5H2O then how many grams it is.
Convert 800. mL to L
800.mL x 1L = 0.8 L
1000mL
0.8 L x 0.6 mol = 0.48 mol
1L
Cu = 63.5 g/mol
S = 32.0 g/mol
O = 16.0 g/mol so 4- O = 64.0 g/mol
H2O = 18.0 g/mol so 5 H2O = 90.0 g/mol
Molecular weight of copper (II) sulfate =
0.48 mol x 249.5 g = 119.76 g
1 mol
Getz 2007

3. 5. You have 20.0 grams of Pb(NO3)2. You put it in water and bring the final
volume to 50.0 mL. What is the Molarity?
20.0 grams
50.0 mL
Need to convert g to mol
Need to convert 50.0 mL
Need to use equation moles/liter
Pb = 207.1 g/mol
2 N = 2(14) = 28.0 g/mol
6- O = 6(16)= 96.0 g/mol
207. 1 + 28.0 + 96.0 = 331.1 g/mol
20.0 g x 1 mol = 0.0604 mol
331.1 g
50.0 mL x 1L = 0.050 L
1000mL
0.0604 mol = 1.208 mol/L = 1.2 mol/L = 1.2 M
0.050 L
6. What is the normality of a 2 M solution of H3PO4?
2 moles of H3PO4 x 3 moles H+ = 6 moles H+ so 6N
1 liter 1 mole of H3PO4 1 liter
7. How many grams of NaOH are in 400.0 mL of a 0.5 M solution of NaOH?
400.0 mL
0.5 M
Need to convert 400.0 mL to L
Need to figure out molecular weight of NaOH
400.0 mL x 1L = 0.4000 L
1000mL
0.4000 L x 0.5 mol = 0.2 mol
1L
Getz 2007

4. Na = 23.0 g/mol
O= 16.0 g/mol
H = 1.0 g/mol
NaOH = 40.0 g/mol
0.2 mol x 40.0 g = 8.0 g
1 mol
8. We need 850.0 mL of a 0.45 M solution of NaHCO 3. How much NaHCO3 do
we weigh?
850.0 mL
0.45 M
Convert 850.0 mL to L
Need to figure out mol
Need to figure out molecular weight of NaHCO3 to get to g of NaHCO3
850.0 mL x 1L = 0.8500L
1000mL
0.8500 L x 0.45 moles = 0.3825moles
1L
Na= 23.0 g/mol
H= 1.0 g/mol
C = 12.0 g/mol
3- O = 48.0 g/mol
NaHCO3= 84.0 g/mol
0.3825 moles x 84.0 g = 32.13 g = 32.1 g or 32 g
1 mol
9. Which solution will have a greater volume?
a. 58.5 g of NaCl in water to make 1000 mL of solution
b. 58.5 g of NaCl added to 1000 mL of water
b will have a greater volume, even though it will not be that much greater
Getz 2007

5. 10. Identify the descriptions in problem 9a and 9b as being either Molarity
or molality.
a. Molarity
b. molality
11. Calculate the Molarity or the molality for the descriptions in 9a and 9b.
You need to do question 10 first; figure out which one is a description of
Molarity and which is molality first. Make sure you use the proper units.
NaCl = 58.5 g/mol
So for 9a it would be 1 M and for 9b it would be 1 m
58.5 g x 1 mol = 1 mol
58.5 g
Getz 2007