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Review Pre-AP Chemistry Unit 13: Solutions and Equilibrium
1. A solution consists of a solute dissolved in a solvent.
2. A solution is a homogeneous mixture. (which means same or evenly distrubuted throughout)
Is this type of mixture a pure substance, a compound, both, or neither? (circle one)
3. Classify the following materials as: solution (S) or heterogeneous mixture (HM) or Neither (N)
_HM_ a. smoky air
_S__ g. salt water (NaCl + H2O)
_S__ b. sugar water (C12H22O11 + H2O)
_HM_ h. milk
_S__ c. vinegar (HC2H3O2 + H2O)
_HM_ i. soda with bubbles
_S__ d. honey
_HM_ j. oil-based paint
_N__ e. pure iodine (I2)
_HM_ k. fog
_N__ f. pure distilled water (H2O)
_S__ l. ethanol in water (C2H5OH + H2O)
4. There is a relatively large amount of solute dissolved in a concentrated / dilute solution. (circle one)
There is much more solvent than solute in a dilute solution.
5. Molarity is the ratio of moles of solute per liter (L) of solution.
A 3.0 molar solution (3.0 M) has 3.0 mole(s) of solute in every _1_ liter(s) of solution.
6. To dilute a solution, you can add more solvent which will change the molarity of the
solution, but will not change the number of moles of solute.
7. A(n) electrolyte is an ionic (or very polar) compound that forms ions in
solution and will conduct electricity.
Nonpolar covalent compounds (like N2 , C8H18 , CCl4 , CO2) are typically nonelectrolytes.
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8. Complete the phrase that describes the types of substances that will generally dissolve in each other:
like dissolves like.
Therefore, polar solvents (like water) can dissolve polar solutes (like alcohols, sugars,
ionic compounds, etc.).
But nonpolar solutes (like fats, oils, hydrocarbons, etc.) will dissolve in nonpolar solvents.
9. A solution is formed by adding methanol, CH3OH, to water. Identify each of the following
intermolecular attractions as breaking or forming in the solution process: (circle answers)
breaking / forming
methanol – methanol
breaking / forming
water – water
breaking / forming
methanol – water
10. In general, the stronger the intermolecular attractions between solute and solvent molecules,
the greater the solubility.
11. In general, the stronger the intermolecular attractions between solute and solute molecules,
the lower the solubility.
12. A tincture of iodine can be used as a disinfectant. This solution is can be made by mixing:
2 g of solid iodine (I2)
3 g of sodium iodide (NaI)
55 g of liquid ethanol (C2H5OH)
40 g of water (H2O)
The solvent in this mixture is best identified as _C_ because it is present in the greatest amount.
A.
B.
C.
D.
I2
NaI
C2H5OH
H2O
13. Three factors that typically cause a solid to dissolve faster in a liquid are:
increase / decrease stirring
increase / decrease heat (OR temperature)
increase / decrease exposed surface area (OR decrease particle size)
14. Gases dissolve faster in liquids at low temperatures and at higher pressures.
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15. Label each solution as saturated, unsaturated, or supersaturated based on the addition of solute:
unsaturated
supersaturated
saturated
16. A solution that contains less solute than can be dissolved at that temperature is unsaturated.
17. If solute crystals are added to a saturated solution, the crystals will…
fall to the bottom undissolved.
18. If the addition of a seed crystal causes dissolved solute to crystallize, the original solution was
supersaturated.
For # 19-23, consider the graph of solubility curves below.
19. Which substance has the lowest solubility at 20oC?
KClO3
20. If 40 g of KCl is dissolved in 100 g of water
at 80oC, then the solution is unsaturated.
21. If 80 g of KNO3 is dissolved in 100 g of water
at 40oC, then the solution is supersaturated.
22. If 80 g of KNO3 is dissolved in 100 g of water
at 50oC, then the solution is saturated.
23. How many grams of NH4Cl can dissolve in 100 g
of water at 70oC? 60 g
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24. When a reaction has reached equilibrium, which one of the following could NEVER be true?
A. there are more products than reactants
B. there are more reactants than products
C. there are no reactants left
D. the reactants are still changing into products
E. the products are changing back into reactants
25. When a reaction has reached equilibrium, which of the following is TRUE?
A. the reaction stops
B. the forward reaction continues
C. the reverse reaction continues
D. BOTH the forward and reverse reactions continue (at equal rates , concentrations are constant)
26. A chemical reaction in is dynamic equilibrium when the relative forward and reverse rates of the
reaction are _____________, and the total amounts of reactant and product are _____________.
R(g)
⇄
P(g)
At time A, the concentrations of R and P are:
constant / equal / zero .
At time B, the forward rate of reaction is:
greater than / equal to / less than
the reverse rate of the reaction.
concentration (M)
27. Two gases were placed in a container and allowed to reach equilibrium shown in the equation above.
The diagram below shows how the concentrations of gases R and P in this system changed over time.
At time C, concentrations of R and P are:
constant / equal / zero .
[R]
[P]
A
B
time (s)
C
28. Circle the letter of each sentence that is TRUE about
the saturated solution pictured here:
A. The total amount of dissolved solute remains constant.
B. The total mass of undissolved crystals remains constant.
C. When the rate of solvation equals the rate of crystallization,
a state of dynamic equilibrium exists.
D. If more solute were added to the container,
the total amount of dissolved solute would increase.
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For #29-34, you may use the following formulas:
M=
mol
L
M 1 V 1 = M 2V 2
29. What is the molarity of a solution that contains 9.50 moles of solute in 3.00 L of solution?
9.50 mol = 3.17 M
3.00 L
30. What is the molarity of a solution containing 116 g NaCl of solute in 775 mL of solution?
116 g NaCl x 1 mol NaCl = 1.98 mol NaCl
58.44 g NaCl
1.98 mol = 2.55 M
0.775 L
31. How many moles of solute are in 3.20 L of a 1.50 M solution of sodium chloride?
3.20 L x 1.50 mol = 4.80 mol
1L
32. How many moles of solute are in 40.0 mL of a 0.615 M solution?
0.0400 L x 0.615 mol = 0.0246 mol
1L
33. How many liters of a 0.510 M solution are contain 0.242 moles of solute?
0.242 mol x
1L
= 0.475 L
0.510 mol
34. What mass of glucose (C6H12O6, 180.18 g/mol) is needed to make 5.00 mL of a 0.200 M solution?
0.00500 L x 0.200 mol x 180.18 g = 0.180 g
1L
1 mol
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For #35-39, you may use the following formulas:
M=
mol
L
M 1 V 1 = M 2V 2
35. How many mL of a 0.150 M NaBr solution are needed to make 100 mL of 0.0500 M NaBr?
(0.150 M)(V1) = (0.0500 M)(100 mL)
V1 = 33.3 mL
36. If 20.0 mL of 12.1 M HCl is used to make a 5.00 L aqueous solution, what is the molarity of the dilute
solution? (hint: choose mL or L, not both)
(12.1 M)(0.0200 L) = (5.00 L)(M2)
M2 = 0.0484 M
Cu(s) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 Ag(s)
37. Solid copper is added to an aqueous solution of silver nitrate shown in the reaction above.
What volume of 1.65 M AgNO3(aq) solution is needed to react completely with 0.250 moles of Cu(s) ?
0.250 mol Cu x 2 mol AgNO3 x 1 L AgNO3 = 0.303 L AgNO3
1 mol Cu
1.65 mol AgNO3
2 Al(s) + 3 ZnCl2(aq)  2 AlCl3(aq) + 3 Zn(s)
38. Solid aluminum is added to an aqueous solution of zinc chloride shown in the reaction above.
What volume of 1.20 M ZnCl2(aq) solution is needed to react completely with 9.44 grams of Al(s) ?
9.44 g Al x 1 mol Al x 3 mol ZnCl2 x 1 L ZnCl2 = 0.437 L ZnCl2
26.98 g Al
2 mol Al
1.20 mol ZnCl2
2 KI(aq) + Pb(NO3)2(aq)  2 KNO3(aq) + PbI2(s)
39. Aqueous solutions of potassium iodide and lead(II) nitrate are mixed to produce solid lead(II) iodide
precipitate shown in the reaction above. If 12.7 mL of 0.425 M KI(aq) solution is added to an
excess of Pb(NO3)2(aq) , what is the maximum mass of PbI2(s) precipitate that could be recovered?
( molar mass of PbI2 = 461.00 g/mol )
0.0127 L KI x 0.425 mol KI x 1 mol PbI2 x 461.00 g PbI2 = 1.25 g PbI2
1 L KI
2 mol KI
1 mol PbI2
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