Chemistry 12 – Solutions – Intro Assignment

Chemistry 12 – Solutions – Intro Assignment
Chapter 7 – Solutions and their Concentrations
1. Define the following terms
a)
solution – homogeneous mixture of two or more substance
b)
solute – substance dissolved in solution – generally the one in the lesser amount
c)
solvent – substance doing the dissolving in solution – generally the one in the greater amount
d)
concentrated solution – solution that contains a large amount of dissolved solute
e)
dilute solution – solution that contains a small amount of dissolved solute
f)
aqueous – dissolved in water
g)
miscible – two liquids that are soluble in each other
h)
immiscible – two liquids that are not soluble in each other
i)
alloy – solution of two or more metals
j)
unsaturated – solution that contains less than the maximum amount of dissolved solute at the given
temperature – a solution that is capable of dissolving more solute
k)
saturated – solution that contains the maximum amount of dissolved solute at a given temperature
l)
supersaturated – a solution that contains more dissolved solute than can normally dissolve at a given
temperature – because of this, the solution will be very unstable.
m)
solubility – the mass of solute that will dissolve in a given amount of solvent at a specific temperature
n)
nonpolar solute/solvent – substance where electrons are shared equally – the charges are evenly
distributed within the molecule
o)
polar solute/solvent – substance with polar bonds arranged asymmetrically – unequal sharing of
electrons within the bonds causing an uneven charge distribution within the molecule – this results in positive
and negative poles or regions within the molecule
p)
like dissolves like – general rule for solubility – the more similar substances are in terms of polarity, the
more likely they are to dissolve in each other – polar solvents dissolve ionic and polar solutes and nonpolar
solvents dissolve nonpolar solutes
q)
concentration – describes the amount of solute dissolved in solution
r)
molarity- method of expressing concentration – mole of solute divided by volume (L) of solution
s)
ppm – parts per million – number of units (grams) of solute per million units (grams) of solution – mass of
solute divided by mass of solution multiply by one million
t)
ppb – parts per billion – number of units (grams) of solute per billion units (grams) of solution – mass of
solute divided by mass of solution multiply by one billion
u)
volumetric flask – pear shaped glass container with a long neck. A line on the neck accurately measures a
fixed volume – used to prepare standard solutions
v)
standard
solution
–
a
solution
where
the
exact
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 1 of 10
concentration
is
known
2. Thought Lab – Matching Solutes and Solvents page 241
Answers to Procedure Questions
(a) non-polar
(b) ionic
(c) non-polar molecule with polar functional groups making it soluble in water
Answers to Analysis Questions
1.
2.
(a) Cobalt(II) chloride and sucrose are both soluble in water.
(b) Ionic and polar solutes dissolve well in polar solvents. Non-polar molecules may also dissolve in polar
solvents, if they have polar functional groups.
(a) Iodine is soluble in kerosene.
(b) Non-polar molecules dissolve in non-polar solvents.
3. 7.1 Section Review page 242 – answer questions 7-9
Section 7.1 Review Answers
Student Textbook page 242
1. Solute and solvent
2. (a) steel, bronze, brass
(b) methanol in water; vinegar
(c) air
Other answers are possible.
3. A homogeneous mixture is a solution. Only one phase can be seen in a homogeneous mixture. and the
mixture has uniform properties throughout various samples.
4. The solution will be clear and colourless. At the bottom of the glass will be 1 g of undissolved sodium
chloride.
5. Divide the filtered solution into three equal portions. One portion should be the control. To each of the other
two portions add a crystal of sodium chloride and stir. If the sodium chloride crystal dissolves, the solution
was unsaturated. If the crystal does not dissolve, the solution was saturated.
6. Soluble indicates that a certain amount of solute will dissolve in a given amount of solvent. Miscible refers to
two liquids that will mix together in any amount. Immiscible refers to liquids that will not dissolve in each
other.
7. Ionic solids dissolve best in polar solvents such as water. Examples are KBr and NaCl.
8. If a crystal of potassium bromate is added to the solution, it will dissolve if the solution is unsaturated and will
not dissolve if the solution is saturated.
9. Liquid A was the pure substance as it left no residue showing that it is composed of only one substance; liquid
B was a solution containing a solute or solutes that did not evaporate but was left behind as a residue.
10. Heat each of the liquids and observe the temperature when boiling. The pure substance will have a constant
boiling point and no residue will remain. If you boil the miscible liquids, the lower boiling point liquid will
vaporize first, after which the boiling point will rise. If you boil the solution containing the solid solute, the
boiling point will be constant, but a solid residue will remain.
11. (a) Oil is non-polar substance that does not dissolve in seawater, a polar solvent. The oil is poisonous to fish,
birds and marine mammals, and coastlines were coated with oily sludge. Oil lowers the insulation value
of fur and feathers, so ducks and sea otters died from hypothermia.
(b) The two principal ways that spilt oil is dispersed is by wave action and evaporation. Prince William Sound
has less wave action than the open sea and relatively small tidal currents, factors that slowed the
dispersal of the oil. Because the seas are relatively cold, the oil was slow to evaporate.
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 2 of 10
4. 7.2 Section Review page 254 – answer questions 1, 2 & 6
Section 7.2 Review Answers
Student Textbook page 254
1. Water molecules surround the ions in sodium chloride such that their positive dipole attracts the chloride
ions and their negative dipole attracts the sodium ions. The breaking of sodium-chloride ionic bonds and the
breaking of intermolecular forces between water molecules are both processes that require energy. The
sodium and chloride separate and each ion is hydrated. The formation of hydrated ions is a process that
releases energy.
2. When water vaporizes, intermolecular attractions are broken. Intramolecular bonds still hold the water
molecules together.
3. (a) More solid dissolves at higher temperatures.
(b) Less gas dissolves at higher temperatures.
4. Salt dissociates into ions in solution, while sugar remains as whole molecules. Because ions are charged
particles they can conduct electricity, whereas neutral molecules cannot.
5. Dissolving a solute in water involves breaking the forces holding solute particles together, and also breaking
the intermolecular forces between water molecules. Breaking forces always requires energy. When water
molecules surround the dissolved particles in a process called hydration, new bonds are formed. The
formation of bonds always releases energy. If heat is released when a solute dissolves in water, the energy
changes for the hydration process is greater than the sum of energies of the bond breaking steps. In an
endothermic dissolution, the energy released upon hydration is less than the energy required to break the
intermolecular bonds of the solute and solvent.
6. (a) Ce2(SO3)3
(b) Least Ce2(SO3)3; most NaNO3
(c) NaCl
(d) 85ºC
(e) 10 g A more accurate value would require a larger, more accurate graph.
7. 130 g solute dissolve at 70°C and 90 g solute dissolve at 55°C. Therefore the mass of solute that will
precipitate out of the solution as it is cooled is (130 g – 90 g) = 40 g. (Answers may vary, depending on
student results.)
8. More mineral deposits would be expected near a thermal spring because hotter water would be expected to
have more solutes dissolved in it. As the solution cools on rising to the surface, the solubility of the minerals
will decrease and the solutes will precipitate out.
5. Practice Problems page 258 – answer questions 2-4
1.
a.
14.2g
× 100 = 3.2%
450mL
b.
31.5g
× 100 = 1.8%
1800mL
c.
1.72g
× 100 = 3%
60mL
2.
1.52g
× 100 = 6.31%
24.1mL
3.
mass = (250 mL)(0.145%) = (250)(0.00145) = 0.36 g
4.
NaCl = 0.86%
mass = (350 mL)(0.86%) = (350)(0.0086) = 3.0 g NaCl
KCl = 0.03%
mass = (350 mL)(0.03%) = (350)(0.0003) = 0.1 g KCl
CaCl2 = 0.033%
mass = (350 mL)(0.033%) = (350)(0.00033) = 0.016 g CaCl2
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 3 of 10
6. Practice problems page 261 – answer questions 5, 8, 9
5.
17 g
× 100 = 26%
65g
8.
mass = (60.5 g)(10.5%) = (60.5 g)(0.105) = 6.35 g
9.
mass = (20 g)(75%) = (20 g)(0.75) = 15 g (20 g if answer uses correct number of significant digits)
7. Practice Problems page 263 – answer questions 12-14
12. A 75% solution would contain 75 mL of ethylene glycol per 100 mL of solution. If you dilute this in a 1:1 ratio,
you would have 75 mL of ethylene glycol per 200 mL of solution.
75mL
× 100 = 38%
200mL
13. volume = (2.5 L)(0.72%) = (2.5 L)(0.0072) = 0.018 L or 18 mL
15mL
× 100
15mL + v
15
0.05 =
15 + v
(0.05)(15 + v ) = 15
5% =
0.75 + 0.05v = 15
14.
0.05v = 15 − 0.75
0.05v = 14.25
v = 285 mL of water
8. Practice problems page 265 – answer questions 16-18
16.
ppm=
16=
mass of DDT
mass of Trout
mass of DDT
2500 g
10
10
Mass of DDT = 0.40 g
17.
ppm=
ppm=
mass of chlorine
mass of soluEon
0.0030 g
1000 g
10
10
= 3 ppm
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 4 of 10
18.
ppm=
ppm=
mass of calcium carbonate
mass of soluEon
0.5 g
1000 g
10
10
= 500 ppm
9. 7.4 Section Review page 276 – answer questions 1-2
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 5 of 10
Molarity Review
1.
Calculate the molarity of the following solutions.
a) 1.00 L of a solution contains 0.260 mol of hydrochloric acid.
n 0.260 mol
C 0
0.260 mol/
mol/L
V
1.00 L
b)
250.0 mL of a solution contains 25.0 g of sodium chloride.
mol
mass
25.0 g
0.428 mol
molar mass 58.44377 g/mol
n 0.428 mol
C 1.71 mol/
mol/L
V
0.2500 L
c)
600.0 mL of a solution contains 1.50 g of calcium carbonate.
mol
mass
1.50 g
0.0150 mol
molar mass 100.0872 g/mol
n 0.0150 mol
C 0.0250 mol/
mol/L
V
0.6000 L
d)
325 mL of a solution contains 10.0 g of chromium(III) nitrate nonahydrate.
mol
mass
10.0 g
0.0250 mol
molar mass 400.14822 g/mol
n 0.0250 mol
C 0.0769 mol/
mol/L
V
0.325 L
e)
50.0 mL of a solution contains 15.6 g of ammonium sulfate.
mol
mass
15.6 g
0.118 mol
molar mass 132.14052 g/mol
n 0.118 mol
C 2.36 mol/
mol/L
V
0.0500 L
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 6 of 10
2.
How would you prepare the following solutions?
a) 1.00 L of 3.00 M ammonium chloride.
mol = CV = (3.00 mol/L)(1.00 L) = 3.00 mol
mass = (mol)(molar mass) = (3.00 mol)(53.49146 g/mol) = 160. g
Dissolve 160. g of solid ammonium chloride in enough water to make 1.00 L of solution.
b)
500.0 mL of 0.250 mol/L mercury(II) nitrate.
mol = CV = (0.250 mol/L)(0.500 L) = 0.125 mol
mass = (mol)(molar mass) = (0.125 mol)(324.5998 g/mol) = 40.6 g
Dissolve 40.6 g of solid mercury(II) nitrate in enough water to make 500.0 mL of
solution.
c)
3
125.0 mL of 0.500 kmol/m barium nitrate.
mol = CV = (0.500 mol/L)(0.1250 L) = 0.0625 mol
mass = (mol)(molar mass) = (0.0625 mol)(261.3398 g/mol) = 16.3 g
Dissolve 16.3 g of solid barium nitrate in enough water to make 125.0 mL of solution.
d)
3
3
250.0 cm of 0.100 mol/dm antimony(III) chloride.
mol = CV = (0.100 mol/dm3)(0.2500 dm3) = 0.0250 mol
mass = (mol)(molar mass) = (0.0250 mol)(228.109 g/mol) = 5.70 g
Dissolve 5.70 g of solid antimony(II) chloride in enough water to make 250.0 cm3 of
solution.
3.
What volume of 2.40 kmol/m3 aluminum chloride can be made from 100.0 g of aluminum
chloride?
mol
mass
100.0 g
0.750 mol
molar mass 133.34054 g/mol
n
0.750 mol
V 0.312 L
C 2.40 mol/L
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 7 of 10
4.
-2
What volume of 2.80 x 10 M sodium fluoride contains 0.150 g of sodium fluoride?
mol
mass
0.150 g
0.00357 mol
molar mass 41.988173 g/mol
n
0.00357 mol
V 0.128 L
C 0.0280 mol/L
5.
3
If 20.0 mL of 0.750 mol/dm hydrobromic acid is diluted to a final volume of 90.0 mL, what is the molar
concentration of the hydrobromic acid in the resulting solution?
C1V1 = C2V2
C2 =
6.
V2
=
.
!"/#$.% #$
.& #
= 0.167 mol/L
What is the molar concentration of the potassium hydroxide solution resulting from the mixture of 50.0 mL of
0.150 M potassium hydroxide and 75.0 mL of 0.250 M potassium hydroxide?
total moles
C=
C=
7.
C1 V1
total volume
=
Ca Va + Cb Vb
Va + Vb
=
0.150 mol/L$0.0500 L$+ 0.250 mol/L$0.0750 L$
0.0500 L+0.0750 L
0.0075 mol + 0.01875 mol
= 0.210 mol/L
0.125 L
3
If one drop (0.0500 mL) of 0.200 kmol/m sodium bromide is added to 100.0 mL of water, what is [NaBr] in the
resulting solution.
C1V1 = C2V2
C2 =
8.
C1 V1
V2
=
0.200 mol/L$0.0000500 L$
0.10005 L
= 1.00 · ()*+ mol/L
The density of pure water at 4°C is 1.00 kg/L. What is the molar concentration of water in pure water?
mol
mass
1000 g
55.5 mol
molar mass 18.01528 g/mol
n 55.5 mol
C 55.
55.5 mol/
mol/L
V
1.00 L
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 8 of 10
9.
3
3
Concentrated nitric acid is 15.4 kmol/m . How would you prepare 2.50 L of 0.375 kmol/m nitric acid?
C1V1 = C2V2
V2 =
C1 V1
C2
=
0.375 mol/L$2.50 L$
15.4 mol/L
= 0.0609 L
Dilute 60.9 mL of 15.4 mol/L nitric acid to a final volume of 2.50 L.
10. If 300.0 mL of solution A contains 25.0 g of potassium chloride and 250.0 mL of solution B contains 60.0 g of
potassium chloride, what is the molar concentration of the potassium chloride solution resulting from the
mixture of solutions A and B?
C=
total moles
total volume
mol
C
mass
85.0 g
1.14 mol
molar mass 74.5513 g/mol
1.14 mol
2
2.07 mol/
mol/L
0.5500 L
*11. Solution A is 0.475 M sodium hydroxide. Solution B also contains sodium hydroxide. When 250.0 mL of solution
A is mixed with 400.0 mL of solution B, the resulting solution is 0.325 M sodium hydroxide. What is the molar
concentration of solution B?
C=
total moles
total volume
=
Ca Va + Cb Vb
Va + Vb
0.325 mol/L =
0.475 ,-.//$0.2500 L$ 0 C1 $0.4000 L$
0.2500 L 0 0.4000 L
0.325 mol/L =
0.11875 mol + 0.4000 LCb $
0.6500 L
0.21125 mol = 0.11875 mol + 0.4000 L(Cb)
0.0925 mol = 0.4000 L(Cb)
Cb = 0.231 mol/L
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 9 of 10
*12. Solution X is 0.135 M sodium chloride. Solution Y also contains sodium chloride. When 55.0 mL of solution X is
mixed with 125 mL of solution Y, the resulting solution is 0.165 M sodium chloride. How many grams of sodium
chloride are contained in 300.0 mL of solution Y?
C=
total moles
total volume
=
Ca Va + Cy Vy
Va + Vy
0.165 mol/L =
0.135 ,-.//$0.0550 L$ 0 3C4 50.125 L$
0.0550 L 0 0.125 L
0.165 mol/L =
0.007425 mol + 0.125 L3Cy 5
0.180 L
0.0297 mol = 0.007425 mol + 0.1250 L(Cy)
0.022275 mol = 0.125 L(Cy)
Cy = 0.1782 mol/L
Y
mol = CV = (0.1782 mol/L)(0.3000 L) = 0.05346 mol
Mass = (mol)(molar mass) = (0.05346 mol)(58.44277 g/mol) = 3.12 g
*13. Solution X is 0.125 M barium nitrate and solution Y is 1.50 M barium nitrate. What volume of solution Y must be
added to 250.0 mL of solution X in order to produce a 0.500 M barium nitrate solution.
C=
total moles
total volume
0.500 mol/L =
0.500 mol/L =
=
Ca Va + Cy Vy
Va + Vy
0.125 ,-.//$0.2500 L$ 0 1.50 mol/L$3V4 5
0.2500 L 0 V4
0.003125 mol + 1.50 mol/L$3V4 5
0.2500 L 0 V4
(0.500 mol/L)(0.2500 L + Vy) = 0.003125 mol + 1.50 mol/L(Vy)
0.125 mol + 0.500 mol/L(Vy) = 0.003125 mol + 1.50 mol/L(Vy)
0.125 mol – 0.003125 mol = 1.50 mol/L(Vy) - 0.500 mol/L(Vy)
0.09375 mol = 1.00 mol/L(Vy)
Vy = 0.0938 L or 93.8 mL
Chemistry 12 – Solutions – Intro Assignment – Answers - Page 10 of 10