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 boilingpoint elevation, or the freezingpoint depression. The freezingpoint depression may be found by measuring the freezing point of the solution.
 E—The potassium ion contribution from the KNO_{3} 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 watersoluble 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 MgCl_{2}/1000 mL)(2 mol Cl^{–}/mol MgCl_{2}) = 0.24 mol Cl^{–}
The number of moles needed = [(0.40 – 0.24) mol Cl^{–}] (1 mol CaCl_{2}/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—
 D—This is a dilution problem. V_{before} = (M_{after}) (V_{after})/(M_{before})
 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 freezingpoint 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—Freezingpoint 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. V_{after} = (M_{before} V_{before})/(M_{after})
 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 KNO_{3}/1000 mL)(l mol K^{+}/1 mol KNO_{3})
= 0.300 mol K^{+}
The potassium ion contribution from K_{3}PO_{4} is:
(700.0 mL)(2.0 mol K_{3}PO_{4}/l000 mL)(3 mol K^{+}/l mol K_{3}PO_{4})
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 Na_{2}SO_{4}/1000 mL) (1 mol SO_{4}^{2–}/1 mol Na_{2}SO_{4}) = 0.0030 mol SO_{4}^{2–}
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 K_{3}PO_{4}/L)(212 g K_{3}PO_{4}/1 mol K_{3}PO_{4}) = 130 g K_{3}PO_{4}
(1000 g H_{2}O)(1 mol H_{2}O/18.0 g H_{2}O = 55.6 mol H_{2}O
The mole fraction may now be determined.
(5.2 mol CH_{3}OH)/(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 MgSO_{4}/1000 mL)(100.0 mL)
(120.4 g MgSO_{4}/mol MgSO_{4})
= 60. g MgSO_{4}
(6.0 M HNO_{3})(0.500 L)(1000 mL/1 L)/ (16.0 M HNO_{3}) = 190 mL
(0.80 mol CCl_{4})(153.8 g CCl_{4} /l mol CCl_{4}) (1 kg/1000 g) = 0.12 kg
m CHCl_{3} = (0.20 mol CHCl_{3})/(0.12 kg) = 1.7 m
A: 3 × 0.10 = 0.30 B, C, and D: 2 × 0.10 = 0.20
(10.0 M HNO_{3} = 50.0 mL)/(4.0 M HNO_{3}) = 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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