Exercises
-
In each pair of aqueous systems, which will have the lower vapor pressure?
- pure water or 1.0 M NaCl
- 1.0 M NaCl or 1.0 M C6H12O6
- 1.0 M \(\ce{CaCl2}\) or 1.0 M (NH4)3PO4
-
In each pair of aqueous systems, which will have the lower vapor pressure?
- 0.50 M Ca(NO3)2 or 1.0 M KBr
- 1.5 M C12H22O11 or 0.75 M Ca(OH)2
- 0.10 M Cu(NO3)2 or pure water
-
In each pair of aqueous systems, which will have the higher boiling point?
- pure water or a 1.0 M NaCl
- 1.0 M NaCl or 1.0 M C6H12O6
- 1.0 M \(\ce{CaCl2}\) or 1.0 M (NH4)3PO4
-
In each pair of aqueous systems, which will have the higher boiling point?
- 0.50 M Ca(NO3)2 or 1.0 M KBr
- 1.5 M C12H22O11 or 0.75 M Ca(OH)2
- 0.10 M Cu(NO3)2 or pure water
-
Estimate the boiling point of each aqueous solution. The boiling point of pure water is 100.0°C.
- 0.50 M NaCl
- 1.5 M Na2SO4
- 2.0 M C6H12O6
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Estimate the freezing point of each aqueous solution. The freezing point of pure water is 0.0°C.
- 0.50 M NaCl
- 1.5 M Na2SO4
- 2.0 M C6H12O6
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Explain why salt (NaCl) is spread on roads and sidewalks to inhibit ice formation in cold weather.
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Salt (NaCl) and calcium chloride (\(\ce{CaCl2}\)) are used widely in some areas to minimize the formation of ice on sidewalks and roads. One of these ionic compounds is better, mole for mole, at inhibiting ice formation. Which is that likely to be? Why?
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What is the osmolarity of each aqueous solution?
- 0.500 M NH2CONH2
- 0.500 M NaBr
- 0.500 M Ca(NO3)2
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What is the osmolarity of each aqueous solution?
- 0.150 M KCl
- 0.450 M (CH3)2CHOH
- 0.500 M Ca3(PO4)2
11. A 1.0 M solution of an unknown soluble salt has an osmolarity of 3.0 osmol. What can you conclude about the salt?
12. A 1.5 M NaCl solution and a 0.75 M Al(NO3)3 solution exist on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and the direction of solvent flow, if any, across the membrane.
-
- 1.0 M NaCl
- 1.0 M NaCl
- 1.0 M (NH4)3PO4
2.
- 1.0 M KBr
- 0.75 M Ca(OH)2
- 0.10 M Cu(NO3)2
3.
- 1.0 M NaCl
- 1.0 M NaCl
- 1.0 M (NH4)3PO4
4.
- 1.0 M KBr
- 0.75 M Ca(OH)2
- 0.10 M Cu(NO3)2
5.
6.
- -1.9°C
- -8.6°C
- -3.8°C
7. NaCl lowers the freezing point of water, so it needs to be colder for the water to freeze.
8. \(\ce{CaCl2}\) splits up into 3 ions while NaCl splits up into 2 ions only. \(\ce{CaCl2}\) will be more effective.
9.
- 0.500 osmol
- 1.000 osmol
- 1.500 osmol
10.
- 0.300 osmol
- 0.450 osmol
- 2.50 osmol
11. It must separate into three ions when it dissolves.
12. Both NaCl and Al(NO3)3 have 3.0 osmol. There will be no net difference in the solvent flow.
9.5: Chemical Equilibrium
Concept Review Exercises
1. What is chemical equilibrium?
2. What does the equilibrium constant tell us?
Answers
1. The rate of the forward reaction equals the rate of the reverse reaction.
2. The ratio of products and reactants when the system is at equilibrium.
EXERCISES
1. If the reaction H2 + I2 ⇌2HI is at equilibrium, do the concentrations of HI, H2, and I2 have to be equal?
2. Do the concentrations at equilibrium depend upon how the equilibrium was reached?
3. What does it mean if the Keq is > 1?
4. What does it mean if the Keq is < 1?
5. Does the equilibrium state depend on the starting concentrations?
6. Write an expression for the equilibrium constant K equation.
a. PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)
b. \(2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)\)
7. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: 3C2H2(g)⟶C6H6(g). Which value of K would make this reaction most useful commercially? Explain your answer.
a. K ≈ 0.01
b. K ≈ 1
c. K ≈ 10.
8. Tell whether the reactants or the products are favored at equilibrium:
a. 2NH3(g) ⇌ N2(g) + 3H2(g) K = 172
b. 2SO3(g) ⇌ 2SO2(g) + O2(g) K = 0.230
c. 2NO(g) + Cl2(g) ⇌ 2NOCl(g) K = 4.6×104
d. N2(g) + O2(g) ⇌ 2NO(g) K = 0.050
AnswerS
1. No, the concentrations are constant but the concentrations do not have to be equal.
2. No.
3. More products than reactants are present at equilibrium.
4. More reactants than products present at equilibrium.
5. No. The equilibrium ratio does not depend on the initial concentrations.
6.
a. \(K=\dfrac{[PCl_3][Cl_2]}{[PCl_5]}\)
b. \(K=\dfrac{[O_2]^3}{[O_3]^2}\)
7. The answer is c. K ≈ 10.
Since \(K=\dfrac{[C_6H_6]^2}{[C_2H_2]^3}\) (K≈10), this means that C6H6 predominates over C2H2. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
8.
a. products
b. reactants
c. products
d. reactants
9.6: Le Chatelier's Principle
Concept Review Exercises
1. Define Le Chatelier’s principle.
2. List the three factors types of changes that can disturb the equilibrium of a system.
Answers
1. Le Chatelier’s principle states that a system at equilibrium is disturbed, it will respond in a way to minimize the disturbance.
2. temperature, change in amount of substance, change in pressure through change in volume
EXERCISES
1. How will each change affect the reaction?
PCl5(g) + heat ⇌PCl3(g) + Cl2(g)
- Addition of PCl5
- Addition of Cl2
- Removal of PCl3
- Increasing temperature
- Decreasing temperature
- Decreasing volume
2. How will each change affect the reaction?
HNO2(aq) ⇌H+(aq) + NO2−(aq)
- Removal of HNO2
- Addition of HCl (i.e. adding more H+)
- Increasing volume
- Decreasing volume
- Removal of NO2−
- Addition of OH− (which will react with and remove H+)
3. How will each change affect the reaction?
CO2(g) + C(s) ⇌2CO(g) ΔH=172.5kJ
- Addition of CO2
- Removal of CO2
- Increasing temperature
- Decreasing temperature
- Increasing volume
- Addition of CO
4. How will each change affect the reaction?
H2(g) + I2(g) ⇌2HI(g) ΔH=−9.48kJ
- Addition of H2
- Removal of H2
- Increasing temperature
- Decreasing temperature
- Increasing volume
- Decreasing volume
AnswerS
1.
- shift right
- shift left
- shift right
- shift right
- shift left
- shift left
2.
- shift left
- shift left
- no effect
- no effect
- shift right
- shift right
3.
- shift right
- shift left
- shift right
- shift left
- shift right
- shift left
4.
- shift right
- shift left
- shift left
- shift right
- no effect
- no effect
9.7: Osmosis and Diffusion
Concept Review Exercises
1. What are some of the features of a semipermeable membrane?
2. What do the prefixes hyper, hypo, and iso mean?
Answers
1. A semipermeable membrane allows some substances to pass through but not others.
2. hyper – higher; hypo – lower; iso - same
EXERCISES
1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
- Which one has a higher concentration?
- Which way will water molecules flow?
- Which volume will increase?
- Which volume will decrease?
- What will happen to the concentration of solution A?
- What will happen to the concentration of solution B?
2. Two solutions with different concentrations and compositions are separated by a semipermeable membrane. The left-hand solution is a .50 M solution of MgSO4, while the right-hand solution contains CaCl2 at a concentration of .40 M. Determine the direction of the flow of solvent, left or right.
3. Given the following situations, wherein two tanks of different solutions are separated by a semipermeable membrane, determine the direction of the flow of solvent (water).
a. Solution A contains a 0.40 M concentration of CaCl2, while Solution B contains a 0.45 M concentration of KI
b. Solution A contains a 1.00 M concentration of NH4Cl, while Solution B contains a 1.00 M concentration of CH2O
4. Cells are placed in a solution and the cells then undergo hemolysis. What can be said about the relative concentrations of solute in the cell and the solution?
5. Describe the relative concentrations inside and outside a red blood cell when crenation occurs.
6. A saltwater fish is placed in a freshwater tank. What will happen to the fish? Describe the flow of water molecules to explain the outcome.
7. What makes up the "head" region of a phospholipid? Is it hydrophobic or hydrophilic?
8. What makes up the "tail" region of a phospholipid? Is it hydrophobic or hydrophilic?
Answers
1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
- Solution B
- A →→ B
- B
- A
- increase
- decrease
2. Water (solvent) flows from left to right.
3.
a. Water flows from Solution B to Solution A.
b. Water flows from Solution B to Solution A.
4. Cells contain fluid with higher concentration than solution outside the cell.
5. Cells contain fluid with a lower concentration than the solution outside the cell.
6. Water molecules will flow from the tank water into the fish because the fish has a higher concentration of salt. If the fish absorbs too much water, it will die.
7. The "head" region is a phosphate group and it is hydrophilic.
8. The "tail" is a hydrocarbon tail and it is hydrophobic.
Contributors
Additional Exercises
-
Calcium nitrate reacts with sodium carbonate to precipitate solid calcium carbonate:
\[Ca(NO_3)_{2(aq)} + Na_2CO_{3(aq)} \rightarrow CaCO_{3(s)} + NaNO_{3(aq)}\]
- Balance the chemical equation.
- How many grams of Na2CO3 are needed to react with 50.0 mL of 0.450 M Ca(NO3)2?
- Assuming that the Na2CO3 has a negligible effect on the volume of the solution, find the osmolarity of the NaNO3 solution remaining after the CaCO3 precipitates from solution.
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The compound HCl reacts with sodium carbonate to generate carbon dioxide gas:
\[HCl_{(aq)} + Na_2CO_{3(aq)} \rightarrow H_2O_{(ℓ)} + CO_{2(g)} + NaCl_{(aq)}\]
- Balance the chemical equation.
- How many grams of Na2CO3 are needed to react with 250.0 mL of 0.755 M HCl?
- Assuming that the Na2CO3 has a negligible effect on the volume of the solution, find the osmolarity of the NaCl solution remaining after the reaction is complete.
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Estimate the freezing point of concentrated aqueous HCl, which is usually sold as a 12 M solution. Assume complete ionization into H+ and Cl− ions.
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Estimate the boiling point of concentrated aqueous H2SO4, which is usually sold as an 18 M solution. Assume complete ionization into H+ and HSO4− ions.
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Seawater can be approximated by a 3.0% m/m solution of NaCl in water. Determine the molarity and osmolarity of seawater. Assume a density of 1.0 g/mL.
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Human blood can be approximated by a 0.90% m/m solution of NaCl in water. Determine the molarity and osmolarity of blood. Assume a density of 1.0 g/mL.
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How much water must be added to 25.0 mL of a 1.00 M NaCl solution to make a resulting solution that has a concentration of 0.250 M?
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Sports drinks like Gatorade are advertised as capable of resupplying the body with electrolytes lost by vigorous exercise. Find a label from a sports drink container and identify the electrolytes it contains. You should be able to identify several simple ionic compounds in the ingredients list.
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Occasionally we hear a sensational news story about people stranded in a lifeboat on the ocean who had to drink their own urine to survive. While distasteful, this act was probably necessary for survival. Why not simply drink the ocean water? (Hint: See Exercise 5 and Exercise 6 above. What would happen if the two solutions in these exercises were on opposite sides of a semipermeable membrane, as we would find in our cell walls?)
Answers
-
- Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaNO3(aq)
- 2.39 g
- 1.80 osmol
2.
a. 2HCl (aq) + Na2CO3(aq) → H2O(ℓ) + CO2(g) + 2NaCl (aq)
b. 10.0 g
c. 1.51 M
-
4. 118°C
-
6. molarity = 0.15 M; osmolarity = 0.31 M
7. 75.0 mL
8. magnesium chloride, calcium chloride (answers may vary)
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The osmotic pressure of seawater is too high. Drinking seawater would cause water to go from inside our cells into the more concentrated seawater, ultimately killing the cells.