Solutions and Colligative Properties: Review Questions for AP Chemistry (page 2)
Review the following concepts if necessary:
- Concentration Units for AP Chemistry
- Electrolytes and Nonelectrolytes for AP Chemistry
- Colligative Properties for AP Chemistry
- Colloids for AP Chemistry
You have 25 minutes to answer the following questions. You will be expected to do your calculations without a calculator. You may use the periodic table at the back of the book. For each question, circle the letter of your choice.
- A solution is prepared by dissolving 1.25 g of an unknown substance in 100.0 mL of water. Which procedure from the following list could be used to determine whether the solute is an electrolyte?
- Measure the specific heat of the solution.
- Measure the volume of the solution.
- Measure the freezing point of the solution.
- Determine the specific heat of the solution.
- Determine the volume of the solute.
- What is the final K+ concentration in a solution made by mixing 300.0 mL of 1.0 M KNO3 and 700.0 mL of 2.0 M K3PO4?
- 1.5 M
- 5.0 M
- 3.0 M
- 2.0 M
- 4.5 M
- Strontium sulfate (SrSO4) will precipitate when a solution of sodium sulfate is added to a strontium nitrate solution. What will be the strontium ion, Sr2+, concentration remaining after 30.0 mL of 0.10 M Na2SO4 solution are added to 70.0 mL of 0.20 M Sr(NO3)2 solution?
- 0.14 M
- 0.15 M
- 0.11 M
- 0.20 M
- 0.030 M
- Which of the following is a strong electrolyte when it is mixed with water?
- A solution with a total chloride ion, Cl–, concentration of 1.0 M is needed. Initially, the solution is 0.30 M in MgCl2. How many moles of solid CaCl2 must be added to 400 mL of the MgCl2 solution to achieve the desired concentration of chloride ion?
- Assuming the volumes are additive, what is the final H+(aq) concentration produced by adding 30.0 mL of 0.50 M HNO3 to 70.0 mL of 1.00 M HCl?
- 0.75 M
- 1.50 M
- 1.25 M
- 0.85 M
- 0.43 M
- The molality of a 1.0-molar ethyl alcohol solution may be determined if which of the following is supplied?
- density of the solution
- van't Hoff factor for ethyl alcohol
- temperature of the solution
- volume of the solution
- solubility of ethyl alcohol
- A solution of chloroform, CHCl3, in carbon tetrachloride, CCl4, is nearly ideal. The vapor pressure of chloroform is 170 mm Hg at 20°C, and the vapor pressure of carbon tetrachloride is 87 mm Hg at this temperature. What is the mole fraction of carbon tetrachloride in the vapor over an equimolar solution of these two liquids?
- To prepare 3.0 L of a 0.20-molar K3PO4 solution (molecular weight 212), a student should follow which of the following procedures?
- The student should weigh 42 g of solute and add sufficient water to obtain a final volume of 3.0 L.
- The student should weigh 42 g of solute and add 3.0 Kg of water.
- The student should weigh 130 g of solute and add sufficient water to obtain a final volume of 3.0 L.
- The student should weigh 42 g of solute and add 3.0 L of water.
- The student should weigh 130 g of solute and add 3.0 L of water.
- A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in this solution?
- Choose the aqueous solution with the highest boiling point.
- 0.10 M HI
- 0.10 M (NH4)3PO4
- 0.20 M C2H5OH
- 0.10 M NH4Cl
- 0.10 M NaI
- How many grams of MgSO4 (molar mass 120.4 g/mol) are in 100.0 mL of a 5.0-molar solution?
- 600 g
- 5.0 g
- 12 g
- 60.0 g
- 120 g
- How many milliliters of concentrated nitric acid (16.0-molar HNO3) are needed to prepare 0.500 L of 6.0-molar HNO3?
- 0.19 mL
- 250 mL
- 375 mL
- 190 mL
- 100 mL
- the molal freezing-point constant, Kf , of the solvent
- the freezing point of the pure solvent and the freezing point of the solution
- the mass of the solute
- the volume of the solvent and the mass of the solute
- the mass of the solvent and the boiling point of the solvent
- the mass of the solvent and the mass of the solute
- no additional information is needed
- A student has a solution with a mole fraction of 0.20 of chloroform (molecular mass 119.4) in carbon tetrachloride (molecular mass 153.8). What is the molality of chloroform in the solution?
- 1.7 m
- 0.17 m
- 0.20 m
- 1.3 m
- 1.20 m
- A solution is 10 percent urea by mass. Which item(s) from the following list are needed to calculate the molarity of this solution?
- the density of the solution
- the density of the solvent
- the molecular weight of urea
- I and III
- I only
- II only
- III only
- I and II
- Which of the following aqueous solutions would have the greatest freezing-point depression?
- 0.10 m (NH4)2SO4
- 0.10 m MnSO4
- 0.10 m NaF
- 0.10 m KCI
- 0.10 m CH3OH
- Which of the following aqueous solutions would have the greatest conductivity?
- 0.2 M NaOH
- 0.2 M RbCl
- 0.2 M NH4NO3
- 0.2 M HNO2
- 0.2 M K3PO4
- An aqueous KNO3 solution is cooled from 75°C to 15°C. Which statement from the following list is true?
- The molarity of the solution does not change.
- The molality of the solution decreases.
- The density of the solution does not change.
- The molality of the solution does not change.
- The mole fraction of the solute increases.
- How many milliliters of water must be added to 50.0 mL of 10.0 M HNO3 to prepare 4.00 M HNO3?
- 50.0 mL
- 125 mL
- 500 mL
- 250 mL
- 75.0 mL
- Pick the pair of substances that will most likely obey Raoult's law.
- CH3CH2CH2CH2COOH(l) and C5H12(l)
- C5H12(l) and H2O(l)
- CH3CH2CH2CH2COOH(l) and H2O(l)
- H3PO4(l) and H2O(l)
- C5Hl2(l) and C6H14(l)
- The best method to isolate pure MgSO4 from an aqueous solution of MgSO4 is
- Evaporate the solution to dryness.
- Titrate the solution.
- Electrolyze the solution.
- Use paper chromatography.
- Filter the solution.
- Pick the conditions that would yield the highest concentration of N2(g) in water.
- partial pressure of gas = 1.0 atm; temperature of water = 25°C
- partial pressure of gas = 0.50 atm; temperature of water = 55°C
- partial pressure of gas = 2.0 atm; temperature of water = 25°C
- partial pressure of gas = 2.0 atm; temperature of water = 85°C
- partial pressure of gas = 1.0 atm; temperature of water = 85°C
When using the freezing-point depression method of determining the molar mass of a nonelectrolyte, what information is needed in addition to the above?
Answers and Explanations
- C—If the solute is an electrolyte, the solution will conduct electricity and the van't Hoff factor, i, will be greater than 1. The choices do not include any conductivity measurements; therefore, the van't Hoff factor would need to be determined. This determination is done by measuring the osmotic pressure, the boiling-point elevation, or the freezing-point depression. The freezingpoint depression may be found by measuring the freezing point of the solution.
- E—The potassium ion contribution from the KNO3 is:
- C—The net ionic equation is:
- B—A (nitrous acid) and D (acetic acid) are weak acids, and E (ammonia) is a weak base. Weak acids and bases are weak electrolytes. C (ethanol) is a nonelectrolyte. Potassium nitrate (B) is a water-soluble ionic compound.
- B—The number of moles of chloride ion needed is: (400 mL)(1.0 mol Cl–/1000 mL) = 0.40 mol Cl–
The initial number of moles of chloride ion in the solution is:
(400 mL)(0.30 mol MgCl2/1000 mL)(2 mol Cl–/mol MgCl2) = 0.24 mol Cl–
The number of moles needed = [(0.40 – 0.24) mol Cl–] (1 mol CaCl2/2 mol Cl–) = 0.080 mol
- D—Both of the acids are strong acids and yield 1 mol of H+ each. Calculate the number of moles of H+ produced by each of the acids. Divide the total number of moles by the final volume.
- A—To calculate the molality of a solution, both the moles of solute and the kilograms of solvent are needed. A liter of solution would contain a known number of moles of solute. To convert this liter to mass, a mass to volume relationship (density) is needed.
- E—The mole fraction may be determined by dividing the vapor pressure of the desired substance by the sum of all the vapor pressures.
- C—To produce a molar solution of any type, the final volume must be the desired volume. This eliminates answers D and E. B involves mass of water instead of volume. A calculation of the required mass will allow a decision between A and B.
- C—A 5.2 molal solution has 5.2 mol of methyl alcohol in one Kg (1000 g) of water. The moles of water are needed.
- B—The boiling point depends on the boilingpoint elevation (a colligative property). All colligative properties depend on the concentration of particles present. C is a nonelectrolyte, thus, the concentration of the particles is 0.20 M. All of the other substances are strong electrolytes. The concentration of particles in each of these may be determined by multiplying the molarity given by the van't Hoff factor. Each of these is calculated as follows:
- D—This is a dilution problem. Vbefore = (Mafter) (Vafter)/(Mbefore)
- D—To calculate the molar mass, the mass of the solute and the moles of the solute are needed. The molality of the solution may be determined from the freezing-point depression, and the freezingpoint depression constant (I and II). If the mass of the solvent is known, the moles of the solute may be calculated from the molality. These moles, along with the mass of the solute, can be used to determine the molar mass.
- A—If the mole fraction of chloroform is 0.20 then the solution has a 0.80 mol fraction of carbon tetrachloride. The moles of chloroform and the kilograms of carbon tetrachloride are needed. If 0.20 mol of chloroform are present, the number of kilograms of carbon tetrachloride is:
- A—To calculate the molarity, the moles of urea and the volume of the solution are needed. Density is an intensive property, so any arbitrary volume of solution may be used. One liter is a convenient volume. Using this volume and the density of the solution, you can calculate the mass of the solution. Ten percent of this is the mass of urea. The mass of urea and the molecular weight of urea give the moles of urea.
- A—Freezing-point depression is a colligative property, which depends on the number of particles present. The solution with the greatest concentration of particles will have the greatest depression. The concentration of particles in E (a nonelectrolyte) is 0.10 m. All other answers are strong electrolytes, and the concentration of particles in these may be calculated by multiplying the concentration by the van't Hoff factor.
- E—The strong electrolyte with the greatest concentration of ions is the best conductor. D is a weak electrolyte, not a strong electrolyte. The number of ions for the strong electrolytes may be found by simply counting the ions: A––2, B––2, C––2, E––4. The best conductor has the greatest value when the molarity is multiplied by the number of ions.
- D—Cooling the solution will change the temperature and the volume of the solution. Volume is important in the calculation of molarity and density. A volume change eliminates answers A and C. The mass and the number of moles are not affected by the temperature change. Mole fraction and molality will not change. This eliminates B and E.
- E—This is a dilution problem. Vafter = (Mbefore Vbefore)/(Mafter)
- E—The two most similar substances will be most likely to be ideal.
- A—Solutions cannot be separated by titrations or filtering. Electrolysis of the solution would produce hydrogen and oxygen gas. Chromatography might achieve a minimal separation.
- C—The solubility of a gas is increased by increasing the partial pressure of the gas, and by lowering the temperature.
(300.0 mL)(1.0 mol KNO3/1000 mL)(l mol K+/1 mol KNO3)
= 0.300 mol K+
The potassium ion contribution from K3PO4 is:
(700.0 mL)(2.0 mol K3PO4/l000 mL)(3 mol K+/l mol K3PO4)
4.20 mol K+
The total potassium is 4.50 mol in a total volume of 1.000 L. Thus, the potassium concentration is 4.50 M.
The sodium sulfate solution contains:
(30.0 mL) (0.10 mol Na2SO4/1000 mL) (1 mol SO42–/1 mol Na2SO4) = 0.0030 mol SO42–
The strontium and sulfate ions react in a 1:1 ratio, so 0.0030 mol of sulfate ion will combine with 0.0030 mol of strontium ion leaving 0.011 mol of strontium in a total volume of 100.0 mL. The final strontium ion concentration is:
(30.0 mL)(0.50 mol H+/1000 mL) + (70.0 mL)(1.00 mol H+/1000 mL) = 0.085 mol H+
87 mm Hg/(87 mm Hg + 170 mm Hg) = 0.34
(3.0 L)(0.20 mol K3PO4/L)(212 g K3PO4/1 mol K3PO4) = 130 g K3PO4
(1000 g H2O)(1 mol H2O/18.0 g H2O = 55.6 mol H2O
The mole fraction may now be determined.
(5.2 mol CH3OH)/(5.2 + 55.6) mol = 0.086
A: 2 × 0.10 = 0.20 B: 4 × 0.10 = 0.40
D: 2 × 0.10 = 0.20 E: 2 × 0.10 = 0.20
(5.0 mole MgSO4/1000 mL)(100.0 mL)
(120.4 g MgSO4/mol MgSO4)
= 60. g MgSO4
(6.0 M HNO3)(0.500 L)(1000 mL/1 L)/ (16.0 M HNO3) = 190 mL
(0.80 mol CCl4)(153.8 g CCl4 /l mol CCl4) (1 kg/1000 g) = 0.12 kg
m CHCl3 = (0.20 mol CHCl3)/(0.12 kg) = 1.7 m
A: 3 × 0.10 = 0.30 B, C, and D: 2 × 0.10 = 0.20
(10.0 M HNO3 = 50.0 mL)/(4.0 M HNO3) = 125 mL
The final volume is 125 mL. Since the original volume was 50.0 mL, an additional 75.0 mL must be added.
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