Transcript Slide 1

EXAMPLE 13.1
Calculating Mass Percent
Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2H6O) and 175 mL of H2O (Assume that the
density of water is 1.00 g/mL.)
Begin by setting up the problem. You are
given the mass of ethanol and the volume of
water and asked to find the mass percent of
the solution.
Given:
27.5 g C2H6O
175 mL H2O
Find: mass percent
Equation and Conversion Factor:
You will need the equation that defines mass
percent and the density of water.
To find the mass percent, substitute into the
equation for mass percent. You need the
mass of the solution, which is simply the
mass of ethanol plus the mass of water. The
mass of water is obtained from the volume of
water by using the density as a conversion
factor.
Solution:
Finally, substitute the correct quantities into
the equation and calculate the mass percent
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.1
Calculating Mass Percent
Continued
SKILLBUILDER 13.1
Calculating Mass Percent
Calculate the mass percent of a sucrose solution containing 11.3 g of sucrose and 412.1 mL of water.
(Assume that the density of water is 1.00 g/mL)
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.11; Problems 43, 44, 45, 46, 47, 48.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
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EXAMPLE 13.2
Using Mass Percent in Calculations
A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of the soft drink solution in milliliters
contains 85.2 g of sucrose? (Assume a density of 1.00 g/mL).
You are given the concentration of sucrose in
a soft drink and a mass of sucrose. You are
asked to find the volume of the soft drink
that contains the given mass of sucrose.
Write the mass percent concentration of
sucrose as a conversion factor, remembering
that percent means per hundred. You will
also need the density to use as a conversion
factor between mass and volume of the soft
drink.
Given:
Convert from g solute (C12H22O11) to g
solution using the mass percent in fractional
form as the conversion factor. Convert to mL
using the density.
Solution Map:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
11.5% C12H22O11 by mass
85.2 g C12H22O11
Find: mL solution (soft drink)
Conversion Factors:
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EXAMPLE 13.2
Using Mass Percent in Calculations
Continued
Follow the solution map to solve the
problem.
SKILLBUILDER 13.2
Solution:
Using Mass Percent in Calculations
How much sucrose (C12H22O11) in grams is contained in 355 mL (12 oz) of the soft drink in Example 13.2?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.12; Problems 49, 50, 51, 52, 53, 54.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.3
Calculating Molarity
Calculate the molarity of a solution made by putting 15.5 g NaCl into a beaker and adding water to make 1.50 L of
NaCl solution.
You are given the mass of sodium chloride
(the solute) and the volume of solution. You
are asked to find the molarity of the solution.
You will need the equation that defines
molarity and the molar mass of NaCl.
To calculate molarity, substitute the correct
values into the equation and compute the
answer. However, you must first convert the
amount of NaCl from grams to moles using
the molar mass of NaCl.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Given:
15.5 g NaCl
1.50 L solution
Find: molarity (M)
Conversion Factors:
Solution:
Copyright ©2009 by Pearson Education, Inc.
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EXAMPLE 13.3
Calculating Molarity
Continued
SKILLBUILDER 13.3
Calculating Molarity
Calculate the molarity of a solution made by putting 55.8 g of NaNO 3 into a beaker and diluting to 2.50 L.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.13; Problems 61, 62, 63, 64, 65, 66.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.4
Using Molarity in Calculations
How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?
You are given the molarity of an NaOH
solution and the number of moles of NaOH.
You are asked to find the volume of solution
that contains the given number of moles.
You will need to use the molarity of the
solution (that is given) as a conversion factor
Given:
The solution map begins with mol NaOH
and shows the conversion to liters of solution
using the molarity.
Solution Map:
Solve the problem by following the solution
map.
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
0.114 M NaOH
1.24 mol NaOH
Find: L solution
Conversion Factor:
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.4
Calculating Molarity
Continued
SKILLBUILDER 13.4
Using Molarity in Calculations
How much of a 0.225 M KCl solution contains 55.8 g of KCl?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.14; Problems 67, 68, 69, 70, 71, 72.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.5
Ion Concentration
Determine the molar concentrations of Na+ and PO43– in a 1.50 M Na3PO4 solution.
You are given the concentration of an ionic
solution and asked to find the concentrations
of the component ions.
Given: 1.50 M Na3PO4
Since a formula unit of Na3PO4 contains
3 Na+ ions (as indicated by the subscript), the
concentration of Na+ is three times the
concentration of Na3PO4. Since the same
formula unit contains one PO43– ion, the
concentration of PO43– is equal to the
concentration of Na3PO4.
Solution:
molarity of Na+ = 3(1.50 M) = 4.50 M
molarity of PO43– = 1.50 M
SKILLBUILDER 13.5
Find: molarity (M) of Na+ and PO43–
Ion Concentration
Determine the molar concentrations of Ca2+ and Cl– in a 0.75 M CaCl2 solution.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 79, 80, 81, 82.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.6
Solution Dilution
To what volume should you dilute 0.100 L of a 15 M NaOH solution to obtain a 1.0 M NaOH solution?
You are given the the initial volume and concentration
of an NaOH solution and a final concentration. You are
asked to find the volume required to dilute the solution
to the given final concentration.
The necessary equation is the solution dilution equation
given previously in this section.
Solve the equation for V2 the volume of the final
solution, and substitute the required quantities to
compute V2. You would make the solution by diluting
0.100 L of the stock solution to a total volume of 1.5
L(V2). The resulting solution will have a concentration
of 1.0 M.
SKILLBUILDER 13.6
V1 = 0.100 L
M1 = 15 M
M2 = 1.0 M
Find:
V2
Equation: M1V1 = M2V2
Given:
Solution:
Solution Dilution
How much 6.0 M NaNO3 solution should be used to make 0.585 L of a 1.2 M NaNO3 solution?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.15; Problems 83, 84, 85, 86, 87, 88, 89, 90.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.7
Solution Stoichiometry
Consider the following precipitation reaction:
How much 0.115 M KI solution in liters is required to completely precipitate the Pb 2+ in 0.104 L of 0.225 M
Pb(NO3)2 solution?
You are given the concentration of a reactant,
KI, in a chemical reaction. You are also
given the volume and concentration of a
second reactant, Pb(NO3)2 You are asked to
find the volume of the first reactant that
completely reacts with the given amount of
the second.
Given:
The conversion factors for this problem are
the molarities of the two solutions expressed
in mol of solute per L of solution and the
stoichiometric relationship (from the
balanced equation) between mol KI and mol
Pb(NO3)2.
Conversion Factors:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
0.115 M KI
0.104 L Pb(NO3)2 solution
0.225 M Pb(NO3)2
Find:
L KI solution
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
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EXAMPLE 13.7
Solution Stoichiometry
Continued
The solution map for this problem is similar
to the solution maps for other stoichiometric
problems. First use the volume and molarity
of Pb(NO3)2 solution to get mol Pb(NO3)2
Then use the stoichiometric coefficients from
the equation to convert mol Pb(NO3)2 to mol
KI. Finally, use mol KI to find L KI solution
Follow the solution map to solve the
problem. Begin with volume of Pb(NO3)2
solution and cancel units to arrive at volume
of KI solution
SKILLBUILDER 13.7
Solution Map:
Solution:
Solution Stoichiometry
How many milliliters of 0.112 M Na2CO3 are necessary to completely react with 27.2 mL of 0.135 M HNO3
according to the following reaction?
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.7
Solution Stoichiometry
Continued
SKILLBUILDER PLUS
A 25.0-mL sample of HNO3 solution requires 35.7 mL of 0.108 M Na2CO3
to completely react
with all of the HNO3 in the solution. What was the concentration of the
HNO3 solution?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.16; Problems 91, 92, 93, 94.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.8
Calculating Molality
Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2H6O2) dissolved in 0.500 kg of water.
You are given the mass of ethylene glycol in
grams and the mass of the solvent in
kilograms. You are asked to find the molality
of the resulting solution.
You will need the equation that defines
molality and the molar mass of ethylene
glycol.
Given:
To calculate molality, simply substitute the
correct values into the equation and compute
the answer. However, you must first convert
the amount of C2H6O2 from grams to moles
using the molar mass of C2H6O2.
Solution:
SKILLBUILDER 13.8
17.2 g C2H6O2
0.500 kg H2O
Find: molality (m)
Equation and Conversion Factor:
Calculating Molality
Calculate the molality (m) of a sucrose (C12H22O11) solution containing 50.4 g sucrose and 0.332 kg of water.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.17; Problems 99, 100, 101, 102.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.9
Freezing Point Depression
Calculate the freezing point of a 1.7 m ethylene glycol solution.
You are given the molality of an aqueous
solution and asked to find the freezing point
depression.
Given:
You will need the freezing point depression
equation given previously in this section.
Equation: ΔTf = m × Kf
To solve this problem, simply substitute the
values into the equation for freezing point
depression and calculate ΔTf
Solution:
Find: ΔTf
The actual freezing point will be the freezing
point of pure water (0.00 °C) – ΔTf
SKILLBUILDER 13.9
1.7 m solution
Freezing point = 0.00 °C – 3.2 °C
= –3.2 °
Freezing Point Depression
Calculate the freezing point of a 2.6 m sucrose solution.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 13.18; Problems 103, 104.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.10
Boiling Point Elevation
Calculate the boiling point of a 1.7 m ethylene glycol solution.
You are given the molality of an aqueous
solution and asked to find the boiling point
depression.
Given:
1.7 m solution
You will need the boiling point elevation
equation given previously in this section.
Equation: ΔTb= m × Kb
To solve this problem, simply substitute the
values into the equation for boiling point
elevation and calculate ΔTb
Solution:
Find: boiling point
The actual boiling point of the solution will
be the boiling point of pure water
(100.00 °C) plus ΔTb
SKILLBUILDER 13.10
Boiling Point Elevation
Calculate the boiling point of a 3.5 m glucose solution.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 105, 106, 107, 108.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.11
Calculating Mass Percent
Find the mass percent concentration of a solution containing 19 g of solute and 158 g of solvent.
Given: 19 g solute
158 g solvent
Find: mass percent
Equation:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.12
Using Mass Percent in Calculations
How much KCl in grams is in 0.337 L of a 5.80% mass percent KCl solution? (Assume that the density of the
solution is 1.05 g/mL.)
Given: 5.80% KCl by mass
0.337 L solution
Find: g KCl
Conversion Factors:
Solution Map:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.13
Calculating Molarity
Calculate the molarity of a KCl solution containing 0.22 mol of KCl in 0.455 L of solution.
Given: 0.22 mol KCl
0.455 L solution
Find: molarity (M)
Equation:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.14
Using Molarity in Calculations
How much KCl in grams is contained in 0.488 L of 1.25 M KCl solution?
Given: 1.25 M KCl
0.488 L solution
Find: g KCl
Conversion Factors:
Solution Map:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.15
Solution Dilution
How much of an 8.0 M HCl solution should be used to make 0.400 L of a 2.7 M HCl solution?
Given: M1 = 8.0 M
M2 = 2.7 M
V2 = 0.400 L
Find: V1
Equation:
M1V1 = M2V2
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.16
Solution Stoichiometry
Consider the following reaction:
How much 0.113 M NaOH solution is required to completely neutralize 1.25 L of 0.228 M HCl solution?
Given: 1.25 L HCl solution
0.228 M HCl
0.113 M NaOH
Find: L NaOH solution
Conversion Factors:
Solution Map:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.16
Solution Stoichiometry
Continued
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.17
Calculating Molality
Calculate the molality of a solution containing 0.183 mol of sucrose dissolved in 1.10 kg of water.
Given: 0.183 mol of sucrose
1.10 kg H2O
Find: molality (m)
Equation:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 13.18
Freezing Point Depression and Boiling Point Elevation
Calculate the freezing point of a 2.5 m aqueous sucrose solution.
Given: 2.5 m solution
Find: ΔTf
Equation:
ΔTf = m × Kf
Solution:
Freezing Point = 0.00 °C – 4.7 °C
= –4.7 °C
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.