Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Unit 10:
Solutions
Student Name: _______________________________________
Class Period: ________
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Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Unit 10 Vocabulary:
1. Aqueous: A solution in which the solvent is water.
2. Colligative Property: A property of a solution that is dependent on
concentration. Examples include boiling point, freezing point, and
vapor pressure.
3. Molality: The concentration of a solution measured in moles of solute
per kilogram of solvent.
4. Molarity: The concentration of a solution measured in moles of solute
per liter of solution.
5. Parts per Million: The concentration of a solution measured in mass
of solute per mass of solution multiplied by one million.
6. Percent by Mass: The concentration of a solution measured in mass of
solute per mass of solution multiplied by one hundred.
7. Percent by Volume: The concentration of a solution measured in
volume of solute per volume of solution multiplied by one hundred.
8. Saturated: Any solution that has the maximum concentration of a
dissolved solute possible in a given quantity of solvent at a given
temperature. A saturated solution is a solution at equilibrium.
9. Solubility: The maximum quantity of a solute that may be dissolved
in a given quantity of solvent at a given temperature to make a
saturated solution.
10. Solute: Any substance that is broken apart by a solvent and kept
separate by the solvent particles.
11. Solution: A homogenous mixture formed when a solute dissolves
into a solvent.
12. Solvent: Any substance that attaches to solute particles, breaks the
solute apart, and then keeps the separated particles apart in solution.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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13. Supersaturated: A solution that has an excess of solute beyond the
solubility point for a given temperature. Excess solute will either
precipitate out, or remain in an unstable dissolved state until the
supersaturated solution is disturbed. This causes the excess solute to
precipitate from solution, leaving the solution saturated.
14. Unsaturated: A solution in which there are solvent particles that
have no attached solute particles, and therefore has the capacity to
take more solute into solution.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Unit 10 Homework Assignments:
Assignment: Date: Due:
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Notes page:
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Mixtures:
A mixture is made by physically combining two (or more)
substances without any type of chemical reaction occurring.
Mixing ionic compounds with water forms aqueous solutions
composed of dissolved ions. The polar water molecules attach to the
ions, separating them from other ions. The polar water molecules
keep the ions separate, holding the ions apart. This property is called
molecule-ion attraction.
Topic: Mixtures
Objective: What is the result of combining two dissimilar substances?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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A solid crystal of NaCl is placed into a beaker of water. Immediately
the polar water molecules attract the charged ions. The partially
positive ends of the water molecules (hydrogen atoms) attract the
negatively charged chloride ions. The partially negative ends of the
water molecules (oxygen atoms) attract the positively charged sodium
ions. When enough water molecules attach to an ion, the natural
kinetic motion of the liquid water molecules will “tear” the ions
from each other. This is the molecule-ion attraction.
Topic: Molecule-Ion Attraction
Objective: How are ions situated within a solution?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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The hydrated (aqueous) ions are kept separated by the motion of the
water molecules that are attracted to them. In this example, four
water molecules around each ion are enough to keep the ions
separated. There are still “free” water molecules in the solution
which could “tear” apart, attract, and keep separate additional solute
ions. This is an example of an unsaturated solution. If more NaCl(s)
were added until all water molecules were attached to ions, the
solution would then become saturated. At the saturation point, any
additional NaCl(s) added would simply sink to bottom of the beaker
without going into solution, or drive other Na+1
or Cl-1
ions out of
solution as a precipitate. The number of water molecules required to
keep ions apart depends on water temperature. The higher the
temperature, the greater the average kinetic energy, and the faster the
water molecules move, so fewer water molecules would be required
to keep the ions apart.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Solubility:
The quantity of a solute that may be added to a given quantity of
solvent at a given temperature and pressure is known as solubility.
Saturated: A saturated solution holds as many dissolved particles as
it has the capacity to hold. Capacity is the maximum “spaces”
available for anything. The chemistry room has 23 desks; this gives
Room 218 the capacity for 23 students unless more desks are brought
in, or if desks are taken out. In a saturated aqueous solution, all the
solvent water molecules (desks) have an ion (student butt) attached to
them, so no more ions (students) could come into the solution (Room
218) and find a place to stay (empty desk). Any additional ions
(students) would need to pass through the solution (room).
Unsaturated: An unsaturated solution holds fewer dissolved
particles as it has the capacity to hold. Room 218 has 23 desks, and
the largest chemistry class (for 2014-2015) has 21 students, so there is
capacity for two more students. In an unsaturated aqueous solution,
there are solvent water molecules (desks) without an ion (student
butt) attached to them, so more ions (students) could come into the
solution (Room 218) and find a place to stay (empty desk).
Topic: Solubility
Objective: How may we know the amount that enters into solution?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Supersaturated: A supersaturated solution is very rare solution in
where the solution holds more dissolved solute than is theoretically
possible. A supersaturated solution is an unstable solution that will
cause the excess solute to leave (precipitate) if disturbed. The
solution (Room 218) has 23 desks (water molecules) that each can
attach to one ion (student butt). If there are more than 23 students
(ions) sitting, whereas some desks have more than one student (ion)
sitting in them, this is over capacity, or supersaturated. The solution
can easily be upset if a disturbance (Murdoch) comes in contact with
the system, and forces the extra ions (students) out of the solution
(Room 218).
Factors affecting solubility:
Three factors affect solubility:
1. Temperature
2. Pressure
3. Nature of solute and solvent
Watch Crash Course Chemistry Solutions - YouTube - 8:20
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Temperature:
1. For solid and liquid solutes, solubility in water increases as water
temperature increases.
2. For gaseous solutes, solubility in water decreases as water
temperature increases.
Effect of temperature on solubility of a solid solute in water:
An example of the effect of temperature on aqueous solubility of
solids is the sugar water needed to form rock candy. A saturated
solution is formed at high temperature. As the saturated sugar
solution cools, the sugar solute becomes less soluble, causing some
water molecules to “jump” off a sugar molecule and assist other water
molecules in holding other sugar molecules apart. This allows the
now “free” sugar molecules to come out of solution, and as more
water molecules are pulled off sugar molecules, more and more sugar
will precipitate from the solution as sugar rock candy.
Topic: Temperature and Solubility
Objective: How does temperature affect the solubility of a solution?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Effect of temperature on solubility of a gaseous solute in water:
An example of the effect of temperature on aqueous solubility of
gases is as easy to find as your favorite carbonated beverage. Take
two identical bottles of the same carbonated beverage, place one in
the refrigerator overnight, and leave the other on the kitchen counter.
When you open both bottles side by side you will see that the colder
beverage will release much less carbonation (bubbles) than will the
warmer beverage. The colder solution allowed more gaseous CO2 to
remain dissolved than did the warmer solution. If you keep observing
both bottles, as the colder beverage warms, it will continue to lose
dissolved CO2 (bubbles) during the warming process, while the
warmer bottle will form fewer bubbles during the same elapsed time.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Pressure:
For solids and liquids, pressure has either no effect or a negligible
effect on solubility.
For gaseous solutes, solubility increases as pressure increases.
i. In commercial soda machines, as water travels through a tube,
flavored syrup is added to the water along with carbon dioxide gas.
Gases don’t have an affinity to form aqueous solutions and require
pressure to force the CO2 gas into the water and syrup solution.
When the aqueous syrup and CO2 mixture is bottled, the gas is
“trapped” within the volume of the container, and equilibrium
between dissolved CO2 molecules and free CO2 molecules quickly
forms. When the container is opened, the pressure decreases,
allowing the CO2 to escape as bubbles. CO2 gas is soluble at high
pressures (sealed container) but nearly insoluble at low pressure
(open container).
Topic: Pressure and Solubility
Objective: How does pressure affect solubility of solutes?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Nature of Solute and Solvent (Like dissolves Like):
Polar solutes dissolve in polar solvents.
o Water is a polar molecule and therefore has partially charged ends
that can attract other polar or ionic solutes. This is why ionic
solutes such as NaCl and polar molecular solutes such as glucose
(sugar) may be dissolved in water.
o Water’s polar structure has little attraction for nonpolar molecular
solutes like oil (remember Hank squishing butter?) Oils will not
mix (are immiscible) with water, and being less dense, will float on
top of the water. This is why water is NOT used to extinguish oil
fires; the water will go under the fire, and float the fire to a new
location. Nonpolar gases such as CO2 and O2 are even harder to
force into aqueous solution because of this.
Nonpolar solutes dissolve in nonpolar solvents.
o Oils will not dissolve in polar water solvent, but oils will easily
dissolve in the nonpolar organic solvent benzene (C6H6). Whereas
water is a great polar solvent, benzene is a good nonpolar solvent.
Acetone is a nonpolar solvent used to remove nonpolar mixtures
such as nail polish.
Topic: Polar Nature of Solvent
Objective: Does the polarity of a solute or solute apply in solutions?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Solubility Regents Practice Problems (ungraded):
1. At room temperature, the solubility of which solute listed below would the most
affected by a change in pressure?
a) Sugar(s)
b) Methanol(l)
c) Sodium nitrate(aq)
d) Carbon dioxide(g)
2. As the pressure on a gas confined above a liquid increases, the solubility of the
gas in the liquid
a) Increases
b) Decreases
c) Is unchanged
3. Sublimated carbon dioxide is most soluble in water under the conditions of
a) Low pressure and low temperature
b) High pressure and low temperature
c) Low pressure and high temperature
d) High pressure and high temperature
4. At which temperature given below could water contain the most dissolved gas
at a pressure of 101.3 kPa?
a) 183˚C
b) 283 K
c) 383 K
d) 483˚C
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Reference Table G: Solubility Curves at Standard Pressure:
1. For Table G, the SLOPE of the curve tells you about the solute!
a. IONIC salts solutes have UPWARDS slopes;
b. GAS solutes (SO2 and NH3) have DOWNWARDS slopes.
2. Interpreting Reference Table G:
a. The Table G x-axis is the temperature in ˚C. If you are given a
temperature in K, you’ll need to convert (K - 273 = ˚C). Note that
the numbers are significant to TWO places, as all numbers have a
decimal (.) at the end. Note also that the interval is to the tens
place (10.˚C), so precision may be interpolated to the ones place
and will still be significant to TWO digits.
b. The Table G y-axis is the solubility in grams of solute per 100.
grams of water. Again, the numbers are significant to TWO places,
as all numbers have a decimal (.) at the end. Note also that the
interval is to the tens place (10. grams), so precision may be
interpolated to the ones place and will still be significant to TWO
digits.
c. Also note the segment of the curve for potassium iodide WAY up
in the extreme upper left corner. Students have missed the KI
curve; be aware of its presence BEFORE you need it!
Topic: Solubility Curves
Objective: How may we use solubility curves to determine solubility?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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3. Table G is only useable at standard pressure (1 atm or 101.3 kPa); if
you have other than standard pressure, you’ll need to infer.
Solubility of a solute in 100. grams of water at various temperatures:
1. Each line on Table G represents a saturated aqueous solution of the
given solute for a given temperature. The farther up the y-axis the
saturation line is at a given temperature, the more soluble the solute
is. To find the saturation point at a given temperature, simply find the
given temperature and then go straight vertically up until you
intersect the solubility line you need, and then traverse level left and
read the solubility on the y-axis. The table is set for grams of
solute/100. grams of H2O; if you have other than 100. g of water you
will need to find the correct proportion.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Determining Molecular Polarity:
Degree of saturation according to Table G:
1. Unsaturated:
2. Saturated:
Topic: Degree of Saturation
Objective: How do we use solubility curves to determine saturation?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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3. Supersaturated:
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Solubility for other than 100. g of water solvent:
Reference Table G may only be directly used in the mass of the
solute is dissolved in 100. grams of water as a solvent. With simple
math, by finding the proportionality of the given amount of solvent to
100. g of water as listed on Table G, we can easily calculate the
solubility at the given mass of water.
If given a question with OTHER than 100. grams of water:
1. For the stated temperature in the question, find the saturation
solubility in grams of solute per 100. grams of water.
2. Place the question’s stated amount of water as a numerator and
100. g of water as the denominator. This will give you a
multiplier.
i. If you had MORE than 100. grams of water in your question,
the multiplier factor will be greater than 1.
ii. If you had LESS than 100. grams of water in your question, the
multiplier factor will be less than 1.
3. Multiply the saturation solubility from Table G by the multiplier
factor from Step #2, and you will have your saturation solubility
for other than 100. grams of water.
Topic: Solubility in other than 100. g
Objective: Will Table G determine solubility not in 100. g of water?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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I. How many grams of KClO3 are required to make a saturated
aqueous solution in 100. grams of water at 30.°C?
a. Find 30.°C on the Table G x-axis, then move straight up to the
KClO3 curve. Move straight over to the y-axis, and read the scale
(12 g). This means that 12 grams of KClO3 are soluble in 100.
grams of water at 30.°C.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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II. How many grams of KClO3 are required to make a saturated
aqueous solution in 50. grams of water at 30.°C?
a. We are now asked for the solubility of KClO3 in 50. grams of
water. Place 50. g of water in the numerator, and 100. g of water
in the denominator, and solve.
50. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
100. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 0.50 multiplier
b. We used Table G to determine that 12 g of KClO3 are soluble in
100. g of water. We can use the multiplier factor with the 12 g of
KClO3 and determine the mass of KClO3 soluble in 50. g of
water.
c. 12 g of KClO3 x 0.50 = 6.0 g of KClO3 soluble in 50. g of water at
30.°C
III. How many grams of KClO3 are required to make a saturated
aqueous solution in 200. grams of water at 30.°C?
a. We are now asked for the solubility of KClO3 in 200. grams of
water. Place 200. g of water in the numerator, and 100. g of water
in the denominator, and solve.
200. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
100. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 2.00 multiplier
b. We used Table G to determine that 12 g of KClO3 are soluble in
100. g of water. We can use the multiplier factor with the 12 g of
KClO3 and determine the mass of KClO3 soluble in 200. g of water.
c. 12 g of KClO3 x 2.00 = 24 g of KClO3 soluble in 200. g of water at
30.°C
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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IV. Let’s say you have to make a saturated NaCl solution using tap
water (10°C) but you only have an empty (355 mL) soda can. Of
course you know that water has a density of very close to 1.0
gram/mL near 10°C, so you can use the 355 mL volume of your
soda can to measure out 355 grams of water. How much NaCl will
be needed to form the saturated solution with the 10°C water?
a. First, start at 10°C on Table G and go up until you find the
solubility curve for NaCl (blue dashed line), and then move left
until you reach the y-axis scale (green dotted line). I’ll call it 37.
However, that 37 g of NaCl is for 100 grams of water! Since you
have 355 grams of water, 355 g needs to go in the numerator, with
the Table G standard of 100. grams of water as the denominator.
355 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
100. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 3.55 multiplier
b. We can now take the interpolated 37 grams of NaCl per 100.
grams of water and multiply that by 3.55 to give us the saturation
point of NaCl in 355 grams of water.
37g
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Table G example questions:
1. What is the solubility of HCl in 100. g of water at 70˚C?
a. Find 70˚C on the x-axis, move straight up to the HCl curve, and
then transit level left to the y-axis.
b. 52 g of HCl are soluble in 100. g of water at 70˚C
2. What type of solution saturation do you have if 20. g of KClO3 are
dissolved in 100. g of water at 40.˚C?
a. Find 40.˚C on the x-axis, move straight up to the KClO3 curve, and
then transit level left to the y-axis.
b. 16 g of KClO3 are soluble in 100. g of water at 40.˚C. As you
were asked what 20. g of KClO3 in solution would be, since 20 is
greater than 16, the solution would be supersaturated.
3. For the question above, how can we make the existing solution
saturated?
a. As calculated using Table G, at 40˚C only 16 g of KClO3 could be
dissolved to make the solution of aqueous KClO3 saturated. With
the stated 20. g of KClO3, it was supersaturated. We can make the
solution saturated if we increase the temperature to around 48˚C,
the point that the curve crosses the 20. g solute/100. g H2O level.
b. What happens in the example if we increase the temperature of the
aqueous solution to 75˚C without adding more solute? When we
find 75˚C on the x-axis and go straight up to the curve, we see that
75˚C crosses the curve at 40. g solute/100. g H2O level. That
would mean that with a capacity of 40. g, and a dissolved solute
amount of 20. g, the aqueous solution of 20. g per 100. g of water
would only be about 50% saturated at 75˚C, or about 50%
unsaturated.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Solubility Regents Practice Problems (ungraded):
1. Solubility data for four different salts aqueous solution at 60˚C are given in the
data table below.
Which given salt is the most soluble at 60C?
a) Salt A b) Salt B c) Salt C d) Salt D
2. According to Reference Table G, which of these substances is most soluble at
60˚C?
a) Sodium chloride
b) Potassium chloride
c) Potassium chlorate
d) Ammonium chloride
3. According to Reference Table G, how many grams of potassium nitrate would
be needed to saturate 200 grams of water at 70˚C?
a) 43 g
b) 86 g
c) 134 g
d) 268 g
4. According to Reference Table G, how does decreasing the average kinetic
energy affect the solubility of NH3 and KCl?
a) The solubility of NH3 increases, and the solubility of KCl increases.
b) The solubility of NH3 increases, and the solubility of KCl decreases.
c) The solubility of NH3 decreases, and the solubility of KCl increases.
d) The solubility of NH3 decreases, and the solubility of KCl decreases.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Student name: _________________________ Class Period: _______
Please carefully remove this page from your packet to hand in.
Solutions Homework
Complete the following the chart: (1 pt. ea.)
For the
solutes
below, in
aqueous
solution:
If temperature of the
solvent is increased,
the solubility of this
solute will:
If the temperature of
the solvent is
decreased, the
solubility of this solute
will:
If the surface area of
this solute is
increased, the
solubility of this
solute will:
NH3(g)
KCl(s)
Complete the following chart: (1 pt. ea.)
Sketch the structures of
the compounds.
Is the solute polar,
nonpolar, or ionic?
Will this solute dissolve in
water or benzene?
CH4
KBr
H2S
Explain why, in terms of molecular polarity, why oils will not easily dissolve in
water. (1 pt.)
Cont’d next page
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Use Reference Table G to answer the following: (1 pt. ea.)
________ 1. As the temperature of water decreases, the solubility of gases will?
________ 2. As temperature of water decreases, the solubility of solids will?
________ 3. Which salt on Table G is the least soluble at 90.˚C?
________ 4. Which gas on Table G is the most soluble at 60.˚C?
________
5. Which salt on Table G shows the least change in solubility between
30.˚C and 70.˚C?
For #6 through #8, determine whether the following solutions are Unsaturated,
Saturated, or SuperSaturated in 100. grams of water.
________ 6. 108 g of KNO3 at 60.˚C
________ 7. 20. g of NH4Cl at 30.˚C
________ 8. 45 g of KCl at 60.˚C
________ 9. What is the solubility of NaNO3 in 100. grams of water at 40.˚C?
________ 10. How much KClO3 may be dissolved in 1000. g of water at 30.˚C?
________ 11. What is the solubility of NaCl in 50. g of water at 70.˚C?
________
12. How many grams of NH4Cl may be added to a solution containing
30. g of NH4Cl to make a saturated solution in 90.˚C H2O?
________
13. A solution has 80. g of KNO3 dissolved in 80.˚C H2O. At what
temperature would the solution become saturated?
________
14. A saturated solution of NaNO3 cools from 343 K to 293 K. If the
solution stays saturated, how many grams of NaNO3 precipitate?
________
15. A saturated solution made in 50. g of water at 303 K of NaNO3 is
left out and fully evaporates. What is the mass of NaNO3 left?
________ 16. At what temperature will KCl and HCl have the same solubility?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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37 g of NaCl x 3.55 = 131.55, or about 130 grams of NaCl are soluble in
355 grams of water at 10°C
Concentration:
The concentration of a solution is a measure of the amount of
substance per unit of volume.
a. When discussing solutions, there are several ways to express
concentration.
1. Grams of solute/100 g of solvent (water):
This is how Reference Table G compares solubility.
2. Molarity (M):
Molarity is the number of moles of solute per liter of solvent.
Molarity is the most common unit of concentration used in the
laboratory.
Topic: Concentration
Objective: How do we express the amount of solute per unit volume?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Determining Molarity Experimentally:
We can experimentally determine the molarity of a solution by
1. Measuring the volume of the liquid solution you have
2. Determine the number of moles of solute in the solution
3. Evaporate all solvent leaving only solute
4. Mass the recovered solute
5. Divide the measured mass of recovered solute by the gram formula
mass of the solute
Topic: Determining Molarity
Objective: How do we work with a single element and compound?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Molarity examples:
1. What is the molarity of a solution containing 2.0 moles of KNO3
dissolved in 4.0 L of water?
a. M = moles of solute/L of solution 2.0 𝑚𝑜𝑙𝑒𝑠 𝐾𝑁𝑂3
4.0 𝐿 𝑤𝑎𝑡𝑒𝑟 = 0.5 M KNO3
2. What is the molarity of a solution containing 2.0 moles of HCl
dissolved in 250. mL of water?
a. As molarity is in units of liters (L) of solution, first convert mL to L.
b. 250 mL x 1 𝐿
1000 𝑚𝐿 = 0.250 L
c. M = moles of solute/L of solution 2.0 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙
0.250 𝐿 𝑤𝑎𝑡𝑒𝑟 = 8.0 M HCl
3. What is the molarity of a solution containing 20. grams of NaOH
dissolved in 2.0 L of water?
a. As molarity is in units of moles of solute, first convert grams of
NaOH to moles of NaOH after finding the gram formula mass for
NaOH.
b. NaOH = (23.0 + 16.0 + 1.0) = 40.0 g / mole NaOH
c. Then convert 20. grams of NaOH to moles of NaOH.
d. 20. g NaOH x 1 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻
40.0 𝑔 𝑁𝑎𝑂𝐻 = 0.5 mole NaOH
e. M = moles of solute/L of solution 0.5 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
2.0 𝐿 𝑤𝑎𝑡𝑒𝑟 = 0.25 M NaOH
KNO3
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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4. What is the molarity of a solution containing 60. grams of NaOH
dissolved in 400. mL of water?
a. As molarity is in units of moles of solute, first convert grams of
NaOH to moles of NaOH.
b. 60. g NaOH x 1 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻
40.0 𝑔 𝑁𝑎𝑂𝐻 = 1.5 mole NaOH
c. As molarity is in units of liters (L) of solution, first convert mL to
L.
d. 400 mL x 1 𝐿
1000 𝑚𝐿 = 0.400 L
e. M = moles of solute/L of solution 1.5 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
0.400 𝐿 𝑤𝑎𝑡𝑒𝑟 = 3.75 or 3.8 M
NaOH
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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The basic equation for molarity of molarity = moles of solute/liter of
solution (Reference Table T) may be rearranged as needed.
i. To make a specific volume of a solution with a desired molarity:
1. If you are given a volume and a molarity, you need to solve for
moles of solute.
2. Moles of solute = M (molarity) x L (volume of solution)
ii. Steps to make your solution:
1. Determine the desired concentration (what is the molarity of the
solution you are trying to make? This is usually told to you in the
lab handout.)
2. Determine how much solution you want to make (how many L or
partial L of solution you need? This, again, is usually told to you
in the lab handout.)
3. Multiply the first two step’s answers. This will give you your
amount of solute in moles.
4. 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x L of solution = moles of solute
5. Calculate the gram formula mass for your required solute, to the
nearest tenth of a gram/mole.
Topic: Calculating Molarity in Lab
Objective: How do we make a solution with a desired molarity?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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6. Multiply the number of moles of solute needed from step 3 by the
gram formula mass of your solute.
7. # of moles calculated x 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑔
1 𝑚𝑜𝑙𝑒 = grams of solute needed
8. Measure your calculated grams of solute.
9. Add your solute to a half-full volumetric flask that matches your
needed volume.
10. When the solute is dissolved, fill the flask to the needed volume.
Wow. Seems like a LOT of work, doesn’t it? Why bother? Who would
want to do all that?
Well,
I (or any other chemistry teacher, lab technician, nurse, doctor,
veterinarian, etc.) have to do ALL of those steps frequently. Creating a
solution to a specific molarity is one of the most commonly done tasks
in chemistry. Once you understand the basics, you will be able to do it
for life.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Examples for calculating molarity:
1. How many grams of NaOH are needed to make 4.0 L of 0.50 M
aqueous NaOH?
a. Determine the number of moles of NaOH needed:
Moles = M x L (0.50 moles/L) x (4.0 L) = 2.0 moles of NaOH
b. Convert moles of solute to grams of solute:
Grams = moles of solute x gram formula mass of solute = (2.0
moles NaOH) x (40.0 g NaOH/mole NaOH) = 80. g of NaOH
c. You can now measure and then add 80. grams of NaOH to a
partially filled 4.0 L volumetric flask and add distilled water to the
4.0 L level.
2. How many grams of KCl are needed to make 500. mL of 0.100 M
aqueous KCl?
a. Molarity is in units of liters (L) of solution, first convert mL to L.
b. 500. mL x 1 𝐿
1000 𝑚𝐿 = 0.500 L
c. Determine the number of moles of KCl needed:
Moles = M x L (0.100 moles/L) x (0.500 L) = 0.0500 moles of
KCl
d. Calculate the gram-formula mass of KCl:
KCl = (39.1 + 35.5) = 74.6 g KCl/mole
e. Convert moles of solute to grams of solute:
Grams = moles of solute x gram formula mass of solute = (0.0500
moles KCl) x (74.6 g KCl/mole KCl) = 3.73 g of KCl
f. You can now measure and then add 3.73 grams of KCl to a partially
filled 500. mL volumetric flask and add distilled water to the 500.
mL level.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Parts-per-million (ppm):
Parts-per-million is a mass-to-mass concentration for determining
impurities or pollutants.
Parts-per-million is used to determine trace amounts of ions in
drinking water. PPM may also be used to measure atmospheric or
industrial air pollutants. For some very toxic materials concentrations
may be expressed in parts-per-billion.
i. What is the ppm concentration of lead ions when 0.0000450 g of
lead is dissolved in 100. grams of water?
Ppm = ( 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ) x 1,000,000 = (
0.0000450 𝑔 𝑜𝑓 𝑃𝑏
100. 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ) x 1,000,000
= 0.45 ppm Pb
Topic: Parts-per-Million
Objective: How would we calculate the concentration in mass?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Ppm examples:
1. A well drilled on property near a landfill was tested before a housing
development was constructed. A 1.000 kg sample of the well water
was found to have 0.00330 g of cadmium, a toxic heavy metal found
in rechargeable batteries.
a. What is the ppm concentration of cadmium in the well water?
Ppm is measured in grams of solution, convert kg to grams:
kg x 1000. 𝑔
1.000 𝑘𝑔 = 1000. grams
Ppm = 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x 1,000,000
0.00330 𝑔 𝑜𝑓 𝐶𝑑
1000. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 x 1,000,000 =
3.30 ppm of cadmium
2. What would be ppm concentration of a solution containing 5.00 g of
solute dissolved in 250. g of water?
For this example you are given the components of the solution, not
the total mass of the solution. We need to add both components
(solute and solvent) together.
5.00 g of solute + 250. g of water = 255 g of solution
Ppm = 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 1,000,000
5.00 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
255 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 1,000,000 =
19607.8 ppm 19,600 ppm
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Percent by Mass:
Percent by Mass = (grams of solute/grams of solution) x 100
i. A 25.0 g sample of aqueous NaCl solution is evaporated and 2.0 g of
NaCl crystals are recovered. What is the percent by mass of the NaCl
in the original solution?
ii. Percent by Mass = 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100 =
2.0 𝑔 𝑜𝑓 𝑁𝑎𝐶𝑙
25.0 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100 =
8.0% NaCl by mass
Percent by Volume:
Percent by Volume = (mL of solute/mL of solution) x 100
i. Percent by volume is often used to describe the concentration of
alcohol. Whiskeys are supposed to be 50% ethanol (EtOH) or more;
50% EtOH(aq) is the point where it remains flammable. A customer
thinking it was too watered down might ask for proof it wasn’t
watered-down, and if it didn’t burn, it was not strong enough. This is
the basis for the term “proof” in comparing distilled spirits.
ii. A 50.0 mL sample of an aqueous ethanol solution is distilled to yield
33.2 mL of EtOH. What percent by volume of EtOH was the original
solution?
Percent by volume = 𝑚𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100
33.2 𝑚𝐿 𝑜𝑓 𝐸𝑡𝑂ℎ
50.0 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100
= 66.4% ethanol by volume
Topic: Percent by Mass & Volume
Objective: Can we calculate concentration as a function or percent?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Concentration Regents Practice Problems (ungraded):
1. What is the molarity of a solution containing 20. grams of NaOH in 500 mL of
water?
a) 1 M
b) 2 M
c) 0.5 M
d) 0.04 M
2. How many moles of solute are in 200 mL of a 1 M solution?
a) 1
b) 0.2
c) 0.8
d) 200
3. What is the concentration of a solution containing 10. moles of copper(II)
nitrate in 5000. mL of water?
a) 1.0 M
b) 2.0 M
c) 5.0 M
d) 0.50M
4. What is the total number of grams of NaI(s) needed to make 1.0 liter of a 0.010
M aqueous solution?
a) 15
b) 1.5
c) 0.15
d) 0.015
5. Using your reference tables, which compound listed would form the most
concentrated aqueous solution?
a) AgBr
b) AgCl
c) AgNO3
d) Ag2CO3
6. What is the concentration of a solution, in parts per million, if 0.02 gram of
Na3PO4 is dissolved in 1000. grams of water?
a) 2 ppm
b) 20 ppm
c) 0.2 ppm
d) 0.02 ppm
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Notes page:
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
Page 43 of 57 Website upload 2015 Lecture notes
Student name: _________________________ Class Period: _______
Please carefully remove this page from your packet to hand in.
Concentrations Homework
You must show ALL work, including numerical setup, units, and correctly rounded
answers. (4 pt. ea.)
1. Calculate the molarities of the following solutions:
a) 2.0 moles of NaCl(s) in 4.0 liters of solution
b) 34 grams of CaSO4(s) in 2000. mL of solution
c) 28 grams of KOH(s) in 100. mL of solution
2. Calculate the mass of solute needed to make the following solutions:
a) 1.0 L of 2.0 M NaOH(aq)
b) 500. mL of 10.0 M HCl(aq)
c) 1.0 L of 5.0 M Ba(NO3)2(aq)
3. 2.5 grams of NaOH(s) were recovered after 50.0 mL of NaOH(aq) completely
evaporated. What was the molarity of the original solution?
4. Calculate the percent by mass of 35.8 g of Na2SO4(aq) in 136.3 grams of
solution.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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5. Calculate the percent by volume of 4.55 mL of ethanol in 7.83 mL of solution.
Parts Per Million: (Show ALL work)
6. A sample is found to contain 0.015 g of Pb+2
in 10 grams of tap water.
Calculate the concentration of the lead in ppm for this sample.
7. The tap water analyzed above came from a site near a landfill. The EPA tested
groundwater samples near the landfill, and found the average groundwater
sample around the landfill contained 2.7 x 10-4
grams of Pb+2
per kilogram of
water. State your conclusion if the lead in the tap water came from the landfill,
and back up the answer with your calculations.
8. A 100.0 mL sample of 8.50 ppm aqueous solution completely evaporates.
What will be the mass of the recovered solute?
9. A manufacturing plant has a maximum allowable airborne particulate limit of
0.100 ppm within the plant clean room. If the clean room contains a total of
450. kg of air, what would be the maximum allowed mass (in grams) of total
airborne particulates allowed in the clean room?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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A property of a solution that is dependent on concentration is known as a
Colligative Property.
Colligative Properties of Solutions: (Electrolytes)
An ionic compound or acid that forms an aqueous solution that
conducts electricity is an electrolyte. Electrolytes have free-moving
ions in solution that allow the electrons to flow through the solution.
Pure (distilled) water will not flow electricity as it contains no
dissolved ions (I personally wouldn’t work with electricity while
standing in a puddle containing distilled water!)
Electrolytes will 100% ionize in water in a reaction that resembles a
decomposition reaction. The ionization reaction is called
dissociation, and is a physical change, not a chemical change. The
more ions a solution forms when it dissociates, the higher the boiling
point and the lower the freezing point of the electrolytic solution.
Topic: Colligative Properties
Objective: What properties are based on the solution concentration?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Colligative Properties Examples: (Electrolytes)
1. NaCl(s) Na+1
(aq) + Cl-1
(aq)
a. One mole of sodium chloride dissociates into one mole of sodium
cations and one mole of chlorine anions (two total moles of ions).
2. CaBr2(s) Ca+2
(aq) + 2 Br-1
(aq)
a. One mole of calcium bromide dissociates into one mole of calcium
cations and two moles of chlorine anions (three total moles of ions).
3. Al(NO3)3(s) Al+3
(aq) + 3 NO3-1
(aq)
a. One mole of aluminum nitrate dissociates into one mole of
aluminum cations and three moles of polyatomic nitrate anions (four
total moles of ions).
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Nonelectrolytes:
Covalently bonded substances (except many acids) do not become
electrolytes when forming aqueous solutions. This includes polar
molecules that dissolve, but not dissociate, in water.
Examples of polar molecules that dissolve but don’t dissociate are
sugars (C6H12O6, C12H22O11), ethylene glycol (CH2OHCH2OH), and
ethanol (C2H5OH). These solutes have less of an effect on the
freezing and boiling points of solutions than do ionic compounds as
they don’t dissociate further.
Nonelectrolyte examples:
1. C12H22O11(s) C12H22O11(aq)
a. One mole of solid sucrose dissolves in water to form one mole of
aqueous sucrose solution. No ions are formed, and no charges exist
to flow electrons.
2. CH2OHCH2OH(l) CH2OHCH2OH(aq)
a. One mole of ethylene glycol dissolves in water to form one mole of
aqueous ethylene glycol solution. No ions are formed, and no
charges exist to flow electrons.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Colligative Properties of Solutions: (Phase change)
1. Freezing Point Depression:
The freezing point temperature of water is depressed (decreased)
when a solute is added to the water.
2. Boiling Point Elevation:
The boiling point temperature of water elevates (increases) when
a solute is added to the water.
3. The higher the concentration of a solute in a solution, the lower the
freezing point (larger depression) and the higher the boiling point
(larger elevation) for the solution.
4. The more particles that a solute dissolves or dissociates into, the
lower the freezing point (larger depression) and the higher the boiling
point (larger elevation) for the solution.
Watch Crash Course Chemistry Water and Solutions video 13:33
Topic: Colligative Properties
Objective: What properties are based on the solution concentration?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
Page 49 of 57 Website upload 2015 Lecture notes
Colligative Properties Examples: (Phase Change)
1. Which of the following aqueous solutions would boil at the highest
temperature?
a) 100 g of NaCl in 125 g of water
b) 100 g of NaCl in 250 g of water
c) 100 g of NaCl in 500 g of water
d) 100 g of NaCl in 1000 g of water
Concentration = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
i. If the amount of solute (NaCl) remains the same, the less solvent (water)
there is would make the solution more concentrated.
ii. As concentration increases, the boiling point elevates. The answer
would then be choice ‘A’.
2. Which of the following aqueous solutions would boil at the lowest temperature?
a) 100 g of NaCl in 125 g of water
b) 100 g of NaCl in 250 g of water
c) 100 g of NaCl in 500 g of water
d) 100 g of NaCl in 1000 g of water
i. Concentration = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
ii. If the amount of solute (NaCl) remains the same, the more solvent (water)
there is would make the solution less concentrated.
iii. As concentration decreases, the boiling point decreases. The answer would
then be choice ‘D’.
3. Which of the following aqueous solutions would freeze at the lowest
temperature?
a) 100 g of NaCl in 125 g of water
b) 100 g of NaCl in 250 g of water
c) 100 g of NaCl in 500 g of water
d) 100 g of NaCl in 1000 g of water
i. Concentration = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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ii. If the amount of solute (NaCl) remains the same, the less solvent (water) there
is would make the solution more concentrated.
iii. As concentration increases, the freezing point depresses. The answer would
then be choice ‘A’.
4. Which of the following aqueous solutions would freeze at the highest
temperature?
a) 100 g of NaCl in 125 g of water
b) 100 g of NaCl in 250 g of water
c) 100 g of NaCl in 500 g of water
d) 100 g of NaCl in 1000 g of water
i. Concentration = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
ii. If the amount of solute (NaCl) remains the same, the more solvent (water)
there is would make the solution less concentrated.
iii. As concentration increases, the freezing point increases. The answer would
then be choice ‘D’.
5. Which of the following aqueous solutions will boil at the highest temperature?
a) 1 mole of KBr in 500 g of water
b) 1 mole of MgF2 in 500 g of water
c) 1 mole of AlCl3 in 500 g of water
d) 1 mole of C6H12O6 in 500 g of water
i. The ionic compound AlCl3 has a total of four atoms, and each dissociates into
a separate ion, so a single mole of AlCl3 makes four moles of ions. The more
particles that a solute forms, the higher the boiling point elevates, so the
answer is ‘C’.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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6. Which of the following aqueous solutions will boil at the lowest temperature?
a) 1 mole of KBr in 500 g of water
b) 1 mole of MgF2 in 500 g of water
c) 1 mole of AlCl3 in 500 g of water
d) 1 mole of C6H12O6 in 500 g of water
i. The ionic compound C6H12O6 has a total of 1 molecule, so a single mole of
C6H12O6 makes one mole of particles. The fewer particles that a solute forms,
the lower the boiling point elevates, so the answer is ‘D’.
7. Which of the following aqueous solutions will freeze at the lowest temperature?
a) 1 mole of KBr in 500 g of water
b) 1 mole of MgF2 in 500 g of water
c) 1 mole of AlCl3 in 500 g of water
d) 1 mole of C6H12O6 in 500 g of water
i. The ionic compound AlCl3 has a total of four atoms, and each dissociates into
a separate ion, so a single mole of AlCl3 makes four moles of ions. The more
particles that a solute forms, the lower the freezing point depresses, so the
answer is ‘C’.
8. Which of the following aqueous solutions will freeze at the highest
temperature?
a) 1 mole of KBr in 500 g of water
b) 1 mole of MgF2 in 500 g of water
c) 1 mole of AlCl3 in 500 g of water
d) 1 mole of C6H12O6 in 500 g of water
i. The ionic compound C6H12O6 has a total of 1 molecule, so a single mole of
C6H12O6 makes one mole of particles. The fewer particles that a solute forms,
the less is the effect on the freezing point, so the answer is ‘D’.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Colligative Properties Regents Practice Problems (ungraded):
1. Which aqueous solution will have the lowest freezing point?
a) 1.0 M NaCl
b) 1.0 M C6H12O6
c) 1.0 M C2H5OH
d) 1.0 M CH3COOH
2. As a solute is added to a solvent, what happens to the freezing point
and the boiling point of the solution?
a) The freezing point decreases and the boiling point decreases.
b) The freezing point decreases and the boiling point increases.
c) The freezing point increases and the boiling point decreases.
d) The freezing point increases and the boiling point increases.
3. Compared to pure water, an aqueous solution of calcium chloride has
a
a) Lower boiling point and a lower freezing point.
b) Lower boiling point and a lower freezing point.
c) Higher boiling point and a lower freezing point.
d) Lower boiling point and a higher freezing point.
4. A student dissolves 1.0 mole of sucrose (C12H22O11) in 1000. grams
of water at 1.0 atmosphere. Compared to the boiling point of pure
water, the boiling point of the resulting solution is
1. 273 K
2. 372 K
3. 373 K
4. 374 K
5. Which 1.0 M aqueous solution would have the lowest freezing point?
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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The chart above is used to determine the effect that adding
ethylene glycol (antifreeze) to your cooling system will have
given the concentration.
Using the chart above, determine:
6. How many quarts of antifreeze are needed to protect a 16-quart
cooling system down to -28F?
7. Based on the bottommost (FREEZE/BOIL PROTECTION CHART)
chart, will aqueous ethylene glycol provide a colligative property
effect against freezing and/or boiling? Explain your answer.
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
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Notes page:
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
Page 55 of 57 Website upload 2015 Lecture notes
Student name: _________________________ Class Period: _______
Please carefully remove this page from your packet to hand in.
Colligative Properties Homework
Ionic compounds (containing BOTH a metal and a nonmetal) are broken apart by water
molecules into individual ions. Only ionic bonds may be broken by dissolving, and any
covalent bonds within a polyatomic ion (Reference Table E) remain intact.
For each of the following compounds determine which ions make up the compound and
how many total ions are contained within the compound. 1 pt. ea.
Ionic compound # of and symbol of
cation
# of an symbol of
anion
Total # of ions in
compound
KHCO3
Fe(C2H3O2)2
Mg(OH)2
CaBr2
Ionic compounds (as described above) form ions, so the resulting aqueous solutions are
known as electrolytes. Covalent molecules (containing only nonmetals) do not form ions,
and therefore do not act as electrolytes. However, for ionic compounds and covalent
compounds, the greater the number of separate aqueous molecules (concentration) in a
solution will have a greater impact on both freezing and boiling point. 1 pt. ea.
Compound Ionic or
Molecular?
Electrolyte of
Nonelectrolyte?
How many
aqueous
particles are
formed per?
Rank the order of
freezing/boiling point
impact (1-less impact;
4-most impact)
BaBr2
LiF
C2H6O
Fe(NO3)3
Cont’d next page
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
Page 56 of 57 Website upload 2015 Lecture notes
Multiple choice questions: Circle the correct answer. 1 pt. ea.
1. Which of the following 1 M solutions will have the highest boiling point?
a) NaCl(aq)
b) C6H12O6(aq)
c) K3PO4(aq)
d) Cu(NO3)2(aq)
2. Which of the following 1 M solutions will have the lowest boiling point?
a) NaCl(aq)
b) C6H12O6(aq)
c) K3PO4(aq)
d) Cu(NO3)2(aq)
3. Which of the following 1 M solutions will have the highest freezing point?
a) NaCl(aq)
b) C6H12O6(aq)
c) K3PO4(aq)
d) Cu(NO3)2(aq)
4. Which of the following 1 M solutions will have the lowest freezing point?
a) NaCl(aq)
b) C6H12O6(aq)
c) K3PO4(aq)
d) Cu(NO3)2(aq)
5. Which of the following aqueous solutions will have the highest boiling point?
a) 500. g of solute in 1.00 kg of solvent
b) 500. g of solute in 0.500 kg of solvent
c) 1000. g of solute in 1.00 kg of solvent
d) 1000. g of solute in 0.500 kg of solvent
6. Which of the following aqueous solutions will have the lowest boiling point?
a) 500. g of solute in 1.00 kg of solvent
b) 500. g of solute in 0.500 kg of solvent
c) 1000. g of solute in 1.00 kg of solvent
d) 1000. g of solute in 0.500 kg of solvent
7. Which of the following aqueous solutions will have the highest freezing point?
a) 500. g of solute in 1.00 kg of solvent
b) 500. g of solute in 0.500 kg of solvent
c) 1000. g of solute in 1.00 kg of solvent
d) 1000. g of solute in 0.500 kg of solvent
8. Which of the following aqueous solutions will have the lowest freezing point?
a) 500. g of solute in 1.00 kg of solvent
b) 500. g of solute in 0.500 kg of solvent
c) 1000. g of solute in 1.00 kg of solvent
d) 1000. g of solute in 0.500 kg of solvent
Unit 10: Solutions-Lecture Regents Chemistry ’14-‘15 Mr. Murdoch
Page 57 of 57 Website upload 2015 Lecture notes
Notes page: