Chapter 12: Physical Properties of Solutions
1.What is the molarity of a solution that is 5.50 % by mass oxalic acid (C2H2O4) and has a density of 1.0244 g/mL?
Ans: 0.626 M
Category: Medium Section: 12.3
2.What is the molality of a 0.142 M Na3PO4(aq) solution that has a density of 1.015 g/mL?
Ans: 0.143 m
Category: Medium Section: 12.3
3.What is the mole fraction of sodium phosphate in a 0.142 M Na3PO4(aq) solution that has a
density of 1.015 g/mL?
Ans: 0.00257
Category: Medium Section: 12.3
4.A solution is prepared by adding 40.3 g of Mg(NO3)2 to 127 g of water. Calculate the mole
fraction and molality of magnesium nitrate in this solution.
Ans: mole fraction = 0.0371; molality = 2.14 m
Category: Medium Section: 12.3
5.What is the concentration of O2(g) in water at 25°C exposed to a partial pressure of oxygen of
325 mmHg? The Henry's law constant for oxygen gas at 25°C is 1.3  10–3 mol/L·atm.
Ans: 5.6  10–4 M
Category: Medium Section: 12.3
6.How many grams of propanol (C3H7OH, 60.10 g/mol) would be needed to make 750 mL of a
solution with an osmotic pressure of 25 atm at 25°C? (R = 0.0821 Latm/Kmol)
Ans: 46 g
Category: Medium Section: 12.6
7.An organic compound was prepared and purified by chromatography. A conductivity measurement showed the compound to be a nonelectrolyte. Elemental analysis gave an empirical formula of CH3O. A 0.127 g sample of the compound was dissolved in water and diluted to 100.0 mL, and the osmotic pressure of that solution was determined to 0.492 atm
at 20°C. Determine the molar mass and molecular formula of the compound. (R =
0.0821 Latm/Kmol)
Ans: 62.1 g/mol, consistent with C2H6O2
Category: Difficult Section: 12.6
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Chapter 12: Physical Properties of Solutions
8.How many liters of ethylene glycol antifreeze (C2H6O2) would you add to your car radiator
containing 15.0 L of water if you needed to protect your engine to –17.8°C? (The density
of ethylene glycol is 1.1 g/mL. For water, Kf = 1.86°C/m.)
Ans: 8.1 L
Category: Difficult Section: 12.6
9.A 100.-mL sample of water is taken from the Pacific Ocean, and the water is allowed to evaporate. The salts that remain (mostly NaCl) have a mass of 3.85 g. Calculate the original
concentration of NaCl, in g per liter, in the water sample.
Ans: 38.5 g/L
Category: Easy Section: 12.3
10The solubility of oxygen in water is about 4.5  10–2 g/L. The water portion of an adult's total
blood supply is about 5 liters. How many grams of oxygen could dissolve in 5 liters of
water?
Ans: 0.2 g
Category: Easy Section: 12.5
11.Explain the following, on the basis of osmosis or osmotic pressure: When sprinkled with sugar, a dish of sliced fruit will form its own juice.
Ans: The water inside the fruit cells passes through cell membranes, trying to dilute the
sugar on the outside.
Category: Medium Section: 12.6
12.Explain the following, on the basis of osmosis or osmotic pressure: In trees and plants water
is drawn from the soil up into the branches and leaves.
Ans: The water passes into the cells, which are more concentrated in solutes, and is
drawn up the tree by osmosis.
Category: Medium Section: 12.6
13.Explain the following, on the basis of osmosis or osmotic pressure: Drinking salt water actually dehydrates our tissues.
Ans: The water passes from cells, trying to dilute the salt water.
Category: Medium Section: 12.6
14.Calculate the mass of solute in the following solution: 50.0 mL of 0.0300 M C12H22O11.
Ans: 0.513 g
Category: Medium Section: 12.3
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Chapter 12: Physical Properties of Solutions
15.When 20.0 grams of an unknown compound are dissolved in 500. grams of benzene, the
freezing point of the resulting solution is 3.77°C. The freezing point of pure benzene is
5.444°C, and the Kf for benzene is 5.12°C/m. What is the molar mass of the unknown
compound?
Ans: 120 g/mol
Category: Difficult Section: 12.6
16.The term "proof" is defined as twice the percent by volume of pure ethanol in solution. A solution that is 95% ethanol is 190 proof. What is the molarity of ethanol in a 92-proof ethanol/water solution? (Given: density of ethanol = 0.80 g/cm3; density of water = 1.0
g/cm3)
Ans: 8.0 M
Category: Difficult Section: 12.3
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