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146 Chemistry: Matter and Change
Solving Problems: A Chemistry Handbook
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 14
Practice Problems
11. Suppose you wished to make 0.879 L of 0.250M silver nitrate
by diluting a stock solution of 0.675M silver nitrate. How many
milliliters of the stock solution would you need to use?
12. If 55.0 mL of a 2.45M stock solution of sucrose is diluted with
water to make 168 mL of sucrose solution, what is the molarity
of the final solution?
Molality The molality (m) of a solution is equal to the number
of moles of solute per kilogram of solvent.
Molality (m) ϭ
Example Problem 14-6
Calculating Molality
What is the molality of a solution that contains 16.3 g of potassium
chloride dissolved in 845 g of water?
Convert the mass of solute to moles.
(16.3 g KCl)
ϭ 0.218 mol KCl
The solvent mass, 845 g, must be expressed in kilograms.
(845 g H 2 O)
ϭ 0.845 kg H 2 O
Substitute the known values into the equation for molality and solve.
Molality (m) ϭ
ϭ
ϭ
ϭ 0.258m
Practice Problems
13. What is the molality of the solution formed by mixing 104 g of
silver nitrate (AgNO 3 ) with 1.75 kg of water?
14. Suppose that 5.25 g of sulfur (S 8 ) is dissolved in 682 g of the
liquid solvent carbon disulfide (CS 2 ). What is the molality of
the sulfur solution?
0.258 mol KCl
ᎏᎏ
kg water
0.218 mol KCl
ᎏᎏ
0.845 kg water
moles of solute
ᎏᎏᎏ
kilogram of solvent
1 kg H 2 O
ᎏᎏ
1000 g H 2 O
1 mol KCl
ᎏᎏ
74.6 g KCl
moles of solute
ᎏᎏᎏ
kilogram of solvent
▲
146 Chemistry: Matter and Change
Solving Problems: A Chemistry Handbook
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 14
Practice Problems
11. Suppose you wished to make 0.879 L of 0.250M silver nitrate
by diluting a stock solution of 0.675M silver nitrate. How many
milliliters of the stock solution would you need to use?
12. If 55.0 mL of a 2.45M stock solution of sucrose is diluted with
water to make 168 mL of sucrose solution, what is the molarity
of the final solution?
Molality The molality (m) of a solution is equal to the number
of moles of solute per kilogram of solvent.
Molality (m) ϭ
Example Problem 14-6
Calculating Molality
What is the molality of a solution that contains 16.3 g of potassium
chloride dissolved in 845 g of water?
Convert the mass of solute to moles.
(16.3 g KCl)
ϭ 0.218 mol KCl
The solvent mass, 845 g, must be expressed in kilograms.
(845 g H 2 O)
ϭ 0.845 kg H 2 O
Substitute the known values into the equation for molality and solve.
Molality (m) ϭ
ϭ
ϭ
ϭ 0.258m
Practice Problems
13. What is the molality of the solution formed by mixing 104 g of
silver nitrate (AgNO 3 ) with 1.75 kg of water?
14. Suppose that 5.25 g of sulfur (S 8 ) is dissolved in 682 g of the
liquid solvent carbon disulfide (CS 2 ). What is the molality of
the sulfur solution?
0.258 mol KCl
ᎏᎏ
kg water
0.218 mol KCl
ᎏᎏ
0.845 kg water
moles of solute
ᎏᎏᎏ
kilogram of solvent
1 kg H 2 O
ᎏᎏ
1000 g H 2 O
1 mol KCl
ᎏᎏ
74.6 g KCl
moles of solute
ᎏᎏᎏ
kilogram of solvent
▲