Honors Chemistry
Chapter 15: Molarity, Molality, Normality,
and Mass Percent Worksheet II
Name ____________________________________
Date _____/_____/_____
Period _____
Molarity = Moles of solute / Liters of Solution (abbreviation = M)
Molality = Moles of solute / Kg of Solvent (abbreviation = m)
Normality = number of equivalent of solute x Molarity of Solution (abbreviation = N)
Mass Percent = mass of solute / mass of solution
1. How many moles of ethyl alcohol, C2H5OH, are present in 65 mL of a 1.5 M solution?
65 mL C2H5OH
1 L C2H5OH
1000 mL C2H5OH
1.5 mole C2H5OH
1 L C2H5OH
= 0.098 moles C2H5OH
2. How many liters of a 6.0 M solution of acetic acid CH3COOH, contain 0.0030 moles of acetic acid?
1 L solution
= 5.0 x 10-4 L solution
6.0 moles CH3COOH
0.0030 moles CH3COOH
3. You want 85 g of KOH. How much of a 3.0 m solution of KOH will provide it?
85 g KOH
1 mole
56.11 g KOH
5.0 x 102 g solvent
+
1 kg solvent
3.0 mole KOH
85 g KOH
=
1000 g solvent
1 kg solvent
= 5.0 x 102 g solvent
590 g solution
4. If you dissolve 0.70 moles of HCl in enough water to prepare 250 mL of solution, what is the molarity
of the solution you have prepared?
M
=
0.70 moles HCl
0.250 L solution
=
2.8 M HCl
5. A solution is prepared by adding 2.0 L of 6.0 M HCl to 500 mL of a 9.0 M HCl solution. What is the
molarity of the new solution? (Remember, the volumes are additive)
2.0 L solution
6.0 moles HCl
1 L solution
0.500 L solution
9.0 moles HCl
1 L solution
= 4.5 moles HCl
12 moles HCl
+
4.5 moles HCl
2.0 L solution
+
0.500 L solution
M
=
= 12 moles HCl
17 moles
2.5 L solution
=
=
17 moles
=
2.5 L solution
6.8 M HCl
6. Convert the following Molarities to Normalities.
a.
2.5 M HCl
=
2.5
N
b.
1.4 M H2SO4
=
2.8
N
c.
1.0 M NaOH
=
1.0
N
d.
0.5 M Ca(OH)2
=
1
N
7. A commonly purchased disinfectant is a 3.0% (by mass) solution of hydrogen peroxide (H2O2) in
water. Assuming the density of the solution is 1.0 g/cm3, calculate the molarity and molality of H2O2.
Mass %
=
3.0 g H2O2
100. g solution
3.0 g H2O2
1 mole H2O2
34.02 g H2O2
100. g solution
M
=
x 100
= 0.088 mole H2O2
1 mL solution
1.0 g solution
0.088 mole H2O2
0.10 L solution
=
1 L solution
1000 mL solution
= 0.10 L solution
0.88 M H2O2
100. g solution - 3.0 g solute = 97 g solvent = 0.097 kg solvent
m
=
0.088 mole H2O2
0.097 kg solvent
=
0.91 m H2O2
8. A solution is made by dissolving 25 g of NaCl in enough water to make 1.0 L of solution. Assume
the density of the solution is 1.0 g/cm3. Calculate the molarity and molality of the solution.
25 g NaCl
M
=
1 mole NaCl
58.44 g NaCl
0.43 mole NaCl
1.0 L solution
1.0 L solution
= 0.43 mole NaCl
=
0.43 M NaCl
1000 mL solution
1 L solution
1.0 g
1 cm3 (or 1 mL)
1.0 x 103 g solution - 25 g
=
M
0.43 m NaCl
=
0.43 mole NaCl
1.0 kg solvent
=
= 1.0 x 103 g solution
1.0 x 103 g solute = 1.0 kg solute
9. What is the mass percent of a solution that contains 152 g of KNO3 in 7.86 L of water?
7.86 L H2O
7860 g H2O
1000 mL H2O
1 L H2O
+
152 g KNO3
152 g KNO3
x 100
8010 g solution
=
=
1.000 g H2O
1 mL
8010 g solution
1.90 %
= 7860 g H2O
10. A solution has been prepared to be 9.00 % (by mass) glucose. For every 100.0 grams of solution
present there are
9.00 g glucose
grams of glucose.
mass %
= 9.00 g glucose
100. g solution
x
100
11. How many grams of KBr are contained in 250. grams of a 6.25 % KBr solution?
6.25 %
=
x
x 100
250. g solution
x = 15.6 g KBr
12. What is the concentration of each type of ion and total concentration of ions in a 0.375 M Ammonium
phosphate solution?
0.375 mole (NH4)3PO4
1 L solution
3 mole (NH4)+1
1 mole (NH4)3PO4
= 1.13 M (NH4)+1
0.375 mole (NH4)3PO4
1 L solution
1 mole (PO4)-3
1 mole (NH4)3PO4
= 0.375 M (PO4)-3
1.13 M (NH4)+1
+
0.375 M (PO4)-3
=
1.51 M ions
13. How many moles of chloride ions are in 1.50 L of 4.15 M zinc chloride?
1.50 L solution
4.15 mole ZnCl2
1 L solution
ANSWERS
1. 0.098 moles of alcohol
2. 5.0 x 10-4 L solution
3. 5.9 x 10 2 g solution
4. 2.8M
5. 6.8 M
6. a. 2.5 N
b. 2.8 N
c. 1.0 N
d. 1 N
2 mole Cl-1
1 mole ZnCl2
= 12.5 mole Cl-1
7. 0.88M, 0.91m
8. 0.43M, 0.43m
9. 1.90 % by mass of KNO3
10. 9.00
11. 15.6 g KBr
12. a. 1.13 M of (NH4)+1 and 0.375 M of (PO4)-3
b. 1.51 M of ions
13. 12.45 moles of Cl-1