Honors Chemistry Name ______
Chapter 15: Molarity, Molality, Normality,
and Mass Percent Worksheet II 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 / 1.5 mole C2H5OH / = 0.098 moles C2H5OH1000 mL C2H5OH / 1 L C2H5OH
2. How many liters of a 6.0 M solution of acetic acid CH3COOH, contain 0.0030 moles of acetic acid?
0.0030 moles CH3COOH / 1 L solution / = 5.0 x 10-4 L solution6.0 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 / 1 kg solvent / 1000 g solvent / = 5.0 x 102 g solvent56.11 g KOH / 3.0 mole KOH / 1 kg solvent
5.0 x 102 g solvent + 85 g KOH = 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 = 2.8 M HCl
0.250 L solution
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 / = 12 moles HCl1 L solution
0.500 L solution / 9.0 moles HCl / = 4.5 moles HCl
1 L solution
12 moles HCl + 4.5 moles HCl = 17 moles
2.0 L solution + 0.500 L solution = 2.5 L solution
M = 17 moles = 6.8 M HCl
2.5 L solution
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 x 100
100. g solution
3.0 g H2O2 / 1 mole H2O2 / = 0.088 mole H2O234.02 g H2O2
100. g solution / 1 mL solution / 1 L solution / = 0.10 L solution
1.0 g solution / 1000 mL solution
M = 0.088 mole H2O2 = 0.88 M H2O2
0.10 L solution
100. g solution - 3.0 g solute = 97 g solvent = 0.097 kg solvent
m = 0.088 mole H2O2 = 0.91 m H2O2
0.097 kg solvent
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 / 1 mole NaCl / = 0.43 mole NaCl58.44 g NaCl
M = 0.43 mole NaCl = 0.43 M NaCl
1.0 L solution
1.0 L solution / 1000 mL solution / 1.0 g / = 1.0 x 103 g solution1 L solution / 1 cm3 (or 1 mL)
1.0 x 103 g solution - 25 g = 1.0 x 103 g solute = 1.0 kg solute
M = 0.43 mole NaCl = 0.43 m NaCl
1.0 kg solvent
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 / 1000 mL H2O / 1.000 g H2O / = 7860 g H2O1 L H2O / 1 mL
7860 g H2O + 152 g KNO3 = 8010 g solution
152 g KNO3 x 100 = 1.90 %
8010 g solution
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 x 100
100. g solution
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 / 3 mole (NH4)+1 / = 1.13 M (NH4)+11 L solution / 1 mole (NH4)3PO4
0.375 mole (NH4)3PO4 / 1 mole (PO4)-3 / = 0.375 M (PO4)-3
1 L solution / 1 mole (NH4)3PO4
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 / 2 mole Cl-1 / = 12.5 mole Cl-11 L solution / 1 mole ZnCl2
ANSWERS
1. 0.098 moles of alcohol 7. 0.88M, 0.91m
2. 5.0 x 10-4 L solution 8. 0.43M, 0.43m
3. 5.9 x 10 2 g solution 9. 1.90 % by mass of KNO3
4. 2.8M 10. 9.00
5. 6.8 M 11. 15.6 g KBr
6. a. 2.5 N 12. a. 1.13 M of (NH4)+1 and 0.375 M of (PO4)-3
b. 2.8 N b. 1.51 M of ions
c. 1.0 N 13. 12.45 moles of Cl-1
d. 1 N