CHAPTER 13 HW (CHEM 110) SOLUTIONS
SOLUTIONS
Dissolving Solids
1.)
2.)
Suppose you dissolve a teaspoon of sugar in a cup of water.
a. What substance is the “solute”?
Sugar
b. What substance is the “solvent”?
Water
c. What is the “solution”?
The sugar and water together
The Lewis structure of water is to the right.
a. What is water’s molecular shape?
H
O
δ−
H
H
δ+
Bent
b. Draw another water molecule beside the first in its correct shape. O
H
δ+
c. On your drawing, place δ+, δ− labels on the appropriate atoms on water’s polar bonds.
d. Which atom in water would be attracted to a Na+ ion? The oxygen atom (it’s partial negative).
e. Which atom in water would be attracted to a Cl– ion? The hydrogen atom (it’s partial positive).
3.)
Draw a microscopic picture of the following aqueous solutions, including both solute and solvent. Be
sure to orient the water molecules correctly. (Note: you don’t need to draw any Lewis structures.)
Solution
KNO3 (aq)
MgCl2 (aq)
Drawing
The δ− end of water (oxygen) should point toward all positive ions. The δ+ end of water (hydrogen) should
point toward all negative ions.
4.)
Are the following statements true or false about what happens when sugar dissolves in water?
a. Chemical bonds within the sugar molecules break.
False (Molecules stay intact)
b. Intermolecular bonds between sugar molecules break.
True
c. Intermolecular bonds form between sugar molecules and water.
True
d. The sugar melts.
False (melting ≠ dissolving)
e. The sugar molecules form covalent bonds with the water molecules.
False
f.
The mass of the solution is the sum of the sugar mass and water mass. True
Page 1
Solubility Trends
5.)
The slogan “Like dissolves like” means polar solutes dissolve in polar solvents and nonpolar solutes
dissolve in nonpolar solvents.
6.)
From the following substances A-D,
H
O
H
H
hydrogen bond
H
C
H
O
C
H
H
H
H
H
H
H
C
H
H H
C
Substance A
C
C
H H
H H
C
C
H H
H
H
C
H H
H
H
C
O
C
H
N
H
C
H
O
C
C
H
H
H
H H
Substance C
Substance B
H
H
C
C
H
H H
O H
H
C
C
C
C
H
H
C
C
C
H
H HH3C H H
C
H H
C
C
H H
C
HH3C H H
C
H H
C
C
C
H H
HH3C H
C
H H
C
C
H
C
H
H
H
Substance D
a. Which would be soluble in water? Briefly explain your answer.
Water is polar, so polar substances are “like” it and are soluble. A and C are both polar, (have polar
bonds like C-O and O-H), so would be soluble. D has a polar area (NH2 area) but there are too many
nonpolar areas, so it is mostly nonpolar.
b. Which would be soluble in hexane (C6H14)? Briefly explain your answer.
Hexane has only carbon and hydrogen atoms so must be nonpolar. It would dissolve other nonpolar
substances, like B and D.
c. Show substance A interacting with water (draw it above) and label the type of intermolecular force.
7.)
Would you expect CH4 to be very soluble in water? Briefly explain.
No. CH4 is a nonpolar molecule and water is a polar solvent. They are not “like” each other, so they
don’t dissolve in each other. To be soluble, CH4 would have to be able to hydrogen bond, and it cannot.
8.)
An oil spill spreads out on the surface of the water, rather than dissolving in the water. Explain why.
The structure of oil (see the lecture notes) has many C-H and C-C bonds, which are nonpolar areas. An
oil molecule can be thought of as mostly nonpolar.
“Like dissolves like.” Water is polar, while oil is not, so they are not soluble in each other. Therefore the
oil spreads out on the surface of water because it doesn’t dissolve.
Page 2
Mass Percent
9.)
An aqueous solution that is 9.0% glucose contains 9.0 grams glucose per 100 grams of solution .
Note: 9.0 has SF, but 100 is exact.
10.) Calculate the mass percent of solute in the following solutions:
a. 2.14 g of potassium chloride dissolved in 12.5 g of water.
m% = !!"#$%&
!!"#$%&"'
= x 100%
!.!" !
!".! !
x 100%
= 14.7%
msoln = mKCl + mw = 2.14 g + 12.5 g = 14.6 g
b. 67.1 g CaCl2 added to 275 g water.
m% = !!"#$%&
!!"#$%&"'
= x 100%
!".! !
!"# !
x 100%
= 19.6%
msoln = mCaCl2 + mH2O = 67.1 g + 275 g = 342 g
11.)
How many grams of KBr are contained in 125 g of a 6.25% KBr solution?
m% = 6.25 % KBr = 125 g solution x !.!" ! !"#
!"" ! !"#$
!.!" ! !"#
!"" ! !"#$
= 7.81 g
12.) Medical biopsies are often preserved in formalin during transport to the laboratory for testing (formalin =
10.0% aqueous solution of formaldehyde, along with buffer). How many grams of formaldehyde are
present in 15.0 g of 10.0% formalin?
10.0 % formalin = 15.0 g formalin x !".! ! !"#$%&'(!!"#
!"" ! !"#$%&'( !"#$%&"'
!".! ! !"#$%&'(!!"#
!"" ! !"#$%&'( !"#$%&"'
= 1.50 g formaldehyde
13.) A hard water sample contains 0.0085% Ca2+ by mass. What mass of this hard water contains 1.2 g Ca2+?
(1.2 g Ca2+ is the recommended daily allowance for 19-24 year olds.)
m% = 0.0085 % Ca2+ = 1.2 g Ca2+ x !.!!"# ! !"!!
!"" ! !"#$
!.!!"# ! !"!!
!"" ! !"#$
= 14,000 g solution (hard water)
Page 3
Parts Per Million
14.) A solution that has 20.5 ppm Pb2+ contains 20.5 grams Pb2+per million (1 x 106) grams of solution.
15.) Anti-cavity toothpaste containing fluoride will have between 1,000-1,500 ppm fluoride (F–). How many
grams of fluoride are present in a pea-sized quantity (0.34 g) of fluoridated toothpaste (1200 ppm)?
1200 ppm = !"## ! !"#$%&'(
! ! !"! ! !"#$%&"'
0.34 g toothpaste (solution) x !"## ! !"#$%&'(
! ! !"! ! !"#$%&"'
= 0.00041 g fluoride (F–) or 4.1 x 10-4 g F–
16.) Food regulations state that the amount of cadmium in fish samples should not exceed 1.00 ppm. Based on
this regulation, what is the maximum ‘acceptable’ mass of cadmium that could be present in a small fish
weighing 2.50 pounds?
2.50 lb. x !"#.! ! !"#!
! !".
x !.!! ! !"#$%&$
! ! !"! ! !"#!
= 0.00113 g cadmium
Introduction to Molarity
17.) An aqueous solution that is 2.50 M glucose contains 2.50 moles
glucose per one liter of solution .
18.) Calculate the molarity of each solution.
a. 3.15 mol NaCl dissolved in enough water to make 5.00 L solution.
M = !"#
!
=
!.!" !"# !"#$
!.!! !
= 0.630 M
b. 0.754 mol KNO3 dissolved in enough water to make 225 mL solution.
V = 225 mL x
M = !"#
!
=
! !
!""" !"
= 0.225 L
!.!"# !"# !"!!
= 3.35 M
!.!!" !
c. 6.789 g of NaCl dissolved in enough water to make 250.0 mL of solution.
m = 6.789 g x
! !"#
!".!! !
= 0.1162 mol NaCl
MM = (22.99) + (35.45) = 58.44 g/mol
V = 250.0 mL x
M = !"#
!
=
! !
!""" !"
= 0.2500 L
!.!!"# !"# !"#$
!.!"## !
= 0.4648 M
Page 4
19.) Arrange these molecular-level representations of four aqueous solutions of the same solute in order from
most to least concentrated. Briefly explain your answer, or show your work.
A = ! !"##$
!.! !
= 4 balls/L
B = ! !"##$
!.! !
C = = 1.5 balls/L
! !"##$
!.! !
= 2 balls/L
D = ! !"##$
!.! !
= 3 balls/L
#Balls are related to moles, because it’s counting. So balls/L is similar to mol/L.
Most concentrated: A > D > C > B Least concentrated.
Molarity as a Conversion Factor
20.) Above each double headed arrow, write what value is needed to interconvert the following quantities
using a 0.075 M NaCl (aq) solution.
Molarity
Molar Mass
Grams NaCl
Moles NaCl
(58.44 g/mol)
Volume Solution
(0.075 mol NaCl / 1 L solution)
21.) How many moles of solute does each solution contain?
a. 0.0125 L of 0.104 M HCl
mol solute = 0.0125 L x
!.!"# !"# !"#
!
= 0.00130 M HCl
b. 27.3 mL of 0.223 M NaOH
mol solute = 27.3 mL x
! !
!""" !"
x
!.!!" !"# !"#$
!
= 0.00609 M NaOH
c. 36.8 mL of 0.501 M HNO3
mol solute = 36.8 mL x
! !
!""" !"
x
!.!"# !"# !"#$
!
= 0.0184 M HNO3
Page 5
22.) What volume of solution contains the following?
a. 0.75 mol CaBr2 in a 0.089 M CaBr2 solution
volume = 0.75 mol CaBr2 x
! !"#$
= 8.4 L
!.!"# !"# !"#$%
b. 0.999 mol NaI in a 1.500 M NaI solution
volume = 0.999 mol NaI x
! !"#$
!.!"" !"# !"#
= 0.666 L
c. 2.07 g NaOH in a 0.1087 M NaOH solution
volume = 2.07 g NaOH x
!"# !"#$
!".!! !
x
! !"#$
!.!"#$ !"# !"#$
= 0.476 L
MMNaOH = (22.99) + (16.00) + (1.01) = 40.00 g/mol
23.) What mass of solute does each solution contain?
a. 17.8 mL of 0.119 M CaCl2
mass solute = 17.8 mL x
! !
!""" !"
x
!.!!" !"# !"!#$
!
x
!!".!" !
!"# !"!#$
= 0.235 g CaCl2
MMCaCl2 = 40.08 + 2(35.45) = 110.98 g/mol
b. 35.4 mL of 0.399 M FeCl3
mass solute = 35.4 mL x
! !
!""" !"
x
!.!"" !"# !"#$%
!
x
!"#.!" !
!"# !"#$%
= 2.29 g FeCl3
MMFeCl3 = 55.85 + 3(35.45) = 162.20 g/mol
24.) How many grams of KNO3 are present in 46.1 mL of a 0.559 M KNO3 aqueous solution?
mass solute = 46.1 mL x
! !
!""" !"
x
!.!!" !"# !"#$
!
x
!"!.!! !
!"# !"#$
= 2.61 g KNO3
MMKNO3 = 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
Page 6
Combined Molarity Problems
25.) An alcoholic iodine solution is prepared by dissolving 5.15 g I2 crystals in enough alcohol to make a
volume of 225 mL.
a. What is the solute in the solution?
I2
b. What is the solvent in the solution?
Alcohol
c. Calculate the molarity of the iodine solution.
MMI2 = 2(126.9) = 253.8 g/mol
V = 225 mL x
! !
!""" !"
5.15 g x
M = = 0.225 L
!"#
!
! !"# !!
!"#.! ! !!
=
= 0.0203 mol I2
!.!"!# !"# !"
!.!!" !
= 0.0902 M
26.) What mass of potassium bromide is contained in 135 mL of a 0.251 M aqueous potassium bromide
solution?
V = 135 mL x
! !
!""" !"
x
!.!"# !"#
! !
x
!!".!! !
! !"#
= 4.03 g
MMKBr = 39.10 + 79.90 = 119.00 g/mol
27.) What mass of (NH4)2SO4 is required to make 1.25 L of a 0.250 M solution of (NH4)2SO4?
1.25 L soln x
!.!"# !"# !!! ! !!!
! ! !"#$
x
!"#.!" ! !!! ! !!!
! !"# !!! ! !!!
= 41.3 g (NH4)2SO4
MM = (2)(14.01) + (8)(1.01) + (32.07) + (4)(16.00) = 132.17 g/mol
Page 7