Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
145
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 14
Practice Problems
9. A solution is made by dissolving 17.0 g of lithium iodide (LiI)
in enough water to make 387 mL of solution. What is the
molarity of the solution?
10. Calculate the molarity of a water solution of CaCl 2 , given that
5.04 L of the solution contains 612 g of CaCl 2 .
Diluting molar solutions Suppose you wished to dilute a stock
solution of known concentration to make a given quantity of solution
of lower concentration? You would need to know the volume of stock
solution to use. To find that information, you can use the equation
M 1 V 1 ϭ M 2 V 2
where M 1 and V 1 are the molarity and volume, respectively, of the
stock solution, and M 2 and V 2 are the molarity and volume, respectively, of the dilute solution. The equation can be solved for any of
the four values, given the other three.
Example Problem 14-5
Diluting a Solution
What volume, in milliliters, of a 1.15M stock solution of potassium
nitrate is needed to make 0.75 L of 0.578M potassium nitrate?
Use the following equation.
M 1 V 1 ϭ M 2 V 2
Rearrange the equation to solve for the volume of stock solution, V 1 ,
and substitute the known values into the equation.
V 1 ϭ V 2
ϭ (0.75L)
ϭ 0.377 L
Convert to milliliters.
(0.377 L)
ϭ 377 mL
Thus, in making the 0.578M solution, 377 mL of the 1.15M stock
solution should be diluted with enough water to make 0.75 L of
solution.
1000 mL
ᎏ
L
0.578M
ᎏ
1.15M
M 2
ᎏ
M 1
▲
Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
145
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 14
Practice Problems
9. A solution is made by dissolving 17.0 g of lithium iodide (LiI)
in enough water to make 387 mL of solution. What is the
molarity of the solution?
10. Calculate the molarity of a water solution of CaCl 2 , given that
5.04 L of the solution contains 612 g of CaCl 2 .
Diluting molar solutions Suppose you wished to dilute a stock
solution of known concentration to make a given quantity of solution
of lower concentration? You would need to know the volume of stock
solution to use. To find that information, you can use the equation
M 1 V 1 ϭ M 2 V 2
where M 1 and V 1 are the molarity and volume, respectively, of the
stock solution, and M 2 and V 2 are the molarity and volume, respectively, of the dilute solution. The equation can be solved for any of
the four values, given the other three.
Example Problem 14-5
Diluting a Solution
What volume, in milliliters, of a 1.15M stock solution of potassium
nitrate is needed to make 0.75 L of 0.578M potassium nitrate?
Use the following equation.
M 1 V 1 ϭ M 2 V 2
Rearrange the equation to solve for the volume of stock solution, V 1 ,
and substitute the known values into the equation.
V 1 ϭ V 2
ϭ (0.75L)
ϭ 0.377 L
Convert to milliliters.
(0.377 L)
ϭ 377 mL
Thus, in making the 0.578M solution, 377 mL of the 1.15M stock
solution should be diluted with enough water to make 0.75 L of
solution.
1000 mL
ᎏ
L
0.578M
ᎏ
1.15M
M 2
ᎏ
M 1
▲