Honors Chemistry
Name ______________________________________
Chapter 15: Solutions Worksheet III
Date _____/_____/_____
1. The temperature (oC) at which HCl and KNO3 have the same solubility is
a. 38
b. 68
c. 74
Period _____
d. 100
2. The maximum amount of KClO 3 that can be dissolved in 25 g of water at 100 oC is
a. 39 g
b. 15 g
c. 40 g
d. 60 g
3. The solubility of KNO3 in 100 g of water at 60 oC is
a. 147 g
b. 80 g
c.
20
d. 113 g
4. The order of decreasing solubility for the substances NaCl, NaNO3, HCl, NH4Cl in 100 g of
water at 40 oC is
a. NaNO3 – HCl – NH4Cl – NaCl
b. NH4Cl – NaCl – NaNO3 – HCl
c. NaCl – NH4Cl – HCl – NaNO3
d. HCl – NaNO3 – NaCl – NH4Cl
5. As temperature increase the solubility of a gas typically ___________________.
a. increases
b. decreases
c. remains the same.
6. As pressure over a solution decreases, the amount of gaseous solute dissolved in solution ________.
a. increases
b. decreases
c. remains the same.
7. As the pressure over a solution increases, the solubility of a solid _______________.
a. increases
b. decreases
c. remains the same
8. What are factors that affect the rate of dissolving?
1. Increasing the surface area of the solute.
2. Agitating a solution (stirring or shaking)
3. Heating a solvent
9. How can Brownian motion be applied to the dissolving process?
The energy absorbed by the random collisions between the solvent and the solute causes the
solute particles to dissolve into solution.
10. How can Henry’s Law be used to explain the solubility of a gas in solution?
The solubility of a gas in a liquid is directly proportional to the partial pressure of that gas on
the surface of the liquid.
11. The molarity is a concentration unit, which relates the moles of a solute per liters of solution. A student adds
58.5 g NaCl to a 1L volumetric flask and fills it up with water to the mark. (NaCl molar mass = 58.5 g/mol).
With respect to this situation it’s correct to say that
a. the concentration of the solution is 1 M.
b. the concentration of the solution is more than 1 M
c. the solution is saturated.
d. the concentration of the solution is less than 1 M.
12. A student has 500 mL 0.5 M HCl (solution 1). 100 mL of a second solution of 0.5 M HCl is added to this
(solution 2). With respect to this new solution it can be stated that
a. it has more solute and less solvent.
b. The concentration stays the same.
c. It has less solute.
d. The concentration decreases.
13. A student has 1 L of a 0.1 M solution. This solution is cooled down and its volume decreased to 998 mL. With
respect to this solution it can be stated that
a. The concentration is more than 0.1 M.
b. The concentration is less than 0.1M.
c. The mass of solute increased.
d. The mass of solvent increased.
14. There are four different solutions with different concentrations and volumes.
Solution 1: 450 mL 0.5 M KMnO4
Solution 2: 500 mL 1.00 M NaI
Solution 3: 100 mL 2.25 M KCl
Solution 4: 450 mL 1.00 M CaCl2
The solution that contains the greatest number of moles of solute is
a. 1
b. 2
c. 3
d. 4
15. From question number 14, which solution produces the most number of moles of ions in solution?
a. 1
b. 2
c. 3
d. 4
16. If you want to prepare 500. mL of a 1.00 M NaOH solution. The amount of solute needed is
a. 20.0 g of NaOH
c. 20.0 g of water
b. 40.0 mL of NaOH
d. 480. mL of water
17. What is the equivalent weight of Ca(OH)2?
a. 24.70 g/eq
b. 74.10 g/eq
c. 148.20 g/eq
18. How many equivalents are in 1 mole of H3PO4?
a. 0.3
b. 1
c. 3
d. 100
d. 37.05 g/eq
e. 30
19. How many mole of equivalents are in 1.00 gram of H2SO4? (MM = 98.09 g/mol)
a. 0.0203
b. 0.0102
c. 0.00510
d. 98.1
20. What is the normality of 12.0 M H2SO4?
a. 3.00 N
b. 6.00 N
c. 12.0 N
21. What is the molarity of 1.0 N HI?
a. 7.8 x 10-3 M
b. 130 M
c. 1.0 M
d. 4.00 N
d. 2.0 M
22. 0.500 moles of sodium chloride is dissolved to make 0.0500 liters of solution.
a. 0.0250 M
b. 10.0 M
c. 0.100 M
d. 400. M
23. 0.500 grams of sodium chloride is dissolved to make 0.0500 liters of solution.
a. 0.171 M
b. 10.0 M
c. 0.100 M
d. 800. M
e. 49.0
e. 24.0 N
24. 0.500 grams of sodium chloride is dissolved to make 0.0500 mL of solution.
a. 171 M
b. 10.0 M
c. 0.0100 M
d. 1.00 M
25. If I add 25.0 mL of water to 125 mL of a 0.150 M NaOH solution, what will the molarity of the diluted solution
be?
a. 0.250 M
b. 0.125 M
c. 0.750 M
d. 1.00 M
26. How much water would I need to add to 500. mL of a 2.4 M KCl solution to make a 1.0 M solution?
a. 0.70 L
b. 4.8 L
c. 210 L
d. 1.00 L
27. I have 345 mL of a 1.5 M NaCl solution. If I boil the water until the volume of the solution is 250 mL, what will
the molarity of the solution be?
a. 0.058 M
b. 4.1 M
c. 2.07 M
d. 3.0 M
28. Explain how you would make 450. mL of a 0.250 M NaOH solution.
I would dissolve 4.50 grams of NaOH in 450 mL of water.
29. To what volume will you have to dilute 30.0 mL of a 12.0 M HCl solution to make a 0.350 M HCl solution?
a. 10.3 L
b. 4.8 L
c. 2.10 L
d. 1.03 L
30. Calculate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.
a. 1.58 m
b. 2.54 m
c. 0.394 m
d. 37.5 m
31. 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of water. What is the
molality?
a. 2.28 m
b. 4.17 m
c. 0.759 m
d. 0.195 m
32. 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?
a. 3.33 m
b. 0.600 m
c. 0.300 m
d. 1.03 m
33. Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water.
a. 0.210 m
b. 0.280 m
c. 0.105 m
d. 0.00210 m
34. 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in1.00 kg of water. Calculate the molality.
a. 0.444 m
b. 4.44 m
c. 0.222 m
d. 2.25 m
35. How many moles of ethyl alcohol, C2H5OH, are present in 65 mL of a 1.5 M solution?
0.065 L C2H5OH
1.5 moles C2H5OH = 0.098 moles C2H5OH
1 L C2H5OH
36. How many liters of a 6.0 M solution of acetic acid CH3COOH, contain 0.0030 moles of acetic acid?
0.0030 moles CH3COOH
1 L CH3COOH
= 5.0 x 10-4 L CH3COOH solution
6.0 moles CH3COOH
37. You want 85 g of KOH. How much of a 3.0 m solution of KOH will provide it?
85 g of KOH
3.0 m KOH =
1 mole KOH
56.11 g KOH
1.5 mole KOH
X kg solvent
= 1.5 mole KOH
= 0.50 kg solvent
5.0 x 10 2 g solvent + 85 g solute = 5.9 x 10 2 g KOH solution
38. A solution is prepared by adding 2.0 L of 6.0 M HCl to 500 mL of a 9.0 M HCl solution. What is
the molarity of the new solution? (Remember, the volumes are additive)
2.0 L HCl =
6.0 M HCl
1 L HCl
= 12 moles HCl
0.5 L HCl =
9.0 moles HCl
1 L HCl
= 5 moles HCl
12 moles HCl + 5 moles HCl = 17 moles HCl
17 moles HCl = 6.8 M
2.5 L
39. Convert the following Molarities to Normalities.
a. 2.5 M HCl x 1 eq = 2.5 N
c. 1.0 M NaOH x 1 eq = 1.0 N
b. 1.4 M H2SO4 x 2 eq = 2.8 N
d. 0.5 M Ca(OH)2 x 2 eq = 1 N
40. A commonly purchased disinfectant is a 3.0% (by mass) solution of hydrogen peroxide (H2O2) in
water. Assuming the density of the solution is 1.00 g/cm3, calculate the molarity and molality of
H2O2.
3.0% =
3.0 g H2O2
3.0 g solute
100. g solution
1 mole H2O2
34.02 g H2O2
100. g solution
0.100 L solution
x 100
= 0.088 mole H2O2
1 mL solution
1.00 g solution
1L
1000 mL
1000 mL solution
1 L solution
= 0.100 L H2O2 solution
1.00 g
1 mL
= 100. g H2O2 solution
M = 0.088 mole H2O2 = 0.88 M H2O2
0.100 L solution
100. g solution
- 3.0 g solute
97 g solvent
m = 0.088 mole H2O2 = 0.91 m H2O2
0.097 kg solvent
41. A solution is made by dissolving 25 g of NaCl in enough water to make 1.000 L of solution.
Assume the density of the solution is 1.000 g/cm3. Calculate the molarity and molality of the
solution.
25 g of NaCl
1 mole NaCl
58.44 g NaCl
= 0.43 moles NaCl
M = 0.43 moles NaCl = 0.43 M NaCl
1.000 L solution
1.000 L solution
1000 mL solution
1 L solution
1000. g solution
- 25 g solute
975 g solvent
m = 0.43 moles NaCl = 0.44 m NaCl
0.975 kg solvent
1.00 g
1 mL
= 1000. g H2O2 solution
42. What is the mass percent of a solution that contains 152 g of KNO3 in 7.86 L of water?
7.86 L H2O
1000 mL H2O 1.00 g H2O
1 L H2O
1 mL H2O
7860 g H2O
+ 152 g KNO3
8012 = 8010 g solution
= 7860 g H2O
152 g of KNO3 x 100 = 1.90 % by mass of KNO3
8010 g solution
43. A solution has been prepared to be 9.00 % (by mass) glucose. For every 100.0 grams of solution
present there are 9.00 grams of glucose.
9.00 % = 9.00 g solute x 100
100. g solution
44. How many grams of KBr are contained in 250. grams of a 6.25 % KBr solution?
6.25 % =
x g solute
250. g solution
x 100
X g solute = 15.6 g KBr
45. What is the concentration of each type of ion and total concentration of ions in a 0.375 M
Ammonium phosphate solution?
0.375 moles (NH4)3PO4
1 L (NH4)3PO4
3 moles (NH4)+1
1 moles (NH4)3PO4
= 1.13 M of (NH4)+1
0.375 moles (NH4)3PO4
1 L (NH4)3PO4
1 moles (PO4)-3
1 moles (NH4)3PO4
= 0.375 M of (PO4)-3
1.13 M of (NH4)+1 + 0.375 M of (PO4)-3 = 1.51 M of ions