CHEM125 – CHAPTER 12 – REGULAR HOMEWORK

CHAPTER 12 REGULAR HOMEWORK

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Answer Explanation Chapter 13 Regular Homework

Question 1

Common commercial acids and bases are aqueous solutions with the following properties: HCl, density 1.19 g/cm3 and 34% solute by mass. Calculate the molarity. (1mL = 1 cm3) Answer:Correct

Explanation

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Question 2

Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO3), density 1.42 g/cm3 and 65% solute by mass. Calculate the molarity. Answer:Correct

Explanation

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Question 3

Common commercial acids and bases are aqueous solutions with the following properties: HCl, density 1.19 g/cm3 and 31% solute by mass. Calculate the molality. Answer:Correct

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Question 4

Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO3), density 1.42 g/cm3 and 67% solute by mass. Calculate the molality. Answer:Correct

Explanation

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Question 5

Common commercial acids and bases are aqueous solutions with the following properties: HCl, density 1.19 g/cm3 and 31% solute by mass. Calculate the mole fraction. Answer:Correct

Explanation

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Question 6

Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO3), density 1.42 g/cm3 and 61% solute by mass. Calculate the mole fraction. Answer:Correct

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Question 7

On the basis of the nature of the intermolecular attractions, arrange water, ethanol (C2H5OH), and chloroform (CHCl3) in order of decreasing solubility in carbon tetrachloride (CCl4). Select one:a. H2O > C2H5OH > CHCl3b. C2H5OH > H2O > CHCl3c. CHCl3 > C2H5OH > H2O Correctd. H2O > CHCl3 > C2H5OHe. CHCl3 > H2O > C2H5OH

Explanation

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Question 8

Predict the relative solubility of each compound in the two solvents on the basis of intermolecular attractions. Is NaCl more soluble in water or in carbon tetrachloride? Select one:a. H2O Correctb. CCl4

Explanation

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Question 9

Is iodine (I2) more soluble in water or in benzene? Select one:a. H2Ob. C6H6 Correct

Explanation

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Question 10

Is ethanol (C2H5OH) more soluble in water or in hexane? Select one:a. H2O Correctb. C6H14

Explanation

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Question 11

Is ethylene glycoL (HOCH2CH2OH) more soluble in ethanol or in benzene? Select one:a. C2H5OH Correctb. C6H6

Explanation

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Question 12

Check the correct statements regarding solubilities in liquids. (The following figure for question (c) shows the solubility of some ionic compounds as a function of temperature.)

CHEM125 - CHAPTER 12 - REGULAR HOMEWORK 1

Select one or more:a. Most gases decrease in solubility as the temperature increases. E.g., oxygen becomes less soluble as the temperature increases making it harder for fish to get sufficient oxygen or opening a cold soda can has less pop than opening a warm one under the same conditions (both shaken or neither shaken). Correctb. For a gas to decrease in solubility with increasing temperature means the process is exothermic. (LeChatelier’s principle–system moves to reduce stress.) Correctc. In the above figure, KNO3 has a steeper slope than NaCl. This means the dissolution of KNO3 is more endothermic than the dissolution of NACl. Correct

Explanation

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Question 13

At 25 oC and 2.0 atm, the enthalpy of solution of neon in water is -2.46 kJ/mol, and its solubility is 9.07 x 10-4 m. Check the conditions for which the solubility of neon is greater than 9.07 x 10-4m. Select one or more:a. 0 oC and 5.0 atm Correctb. 25 oC and 1.00 atmc. 10 oC and 2.0 atm Correctd. 50 oC and 2.0 atme. 50 oC and 1.5 atm

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Question 14

The enthalpy of solution of nitrous oxide (N2O) in water is -12 kJ/mol and its solubility at 20 oC and 1.00 atm is 0.121 g per 100. g of water. Calculate the molal solubility of nitrous oxide in water at 0.800 atm and 20 oC. Hint, first find Henry’s law constant at 20 oC and 1.00 atm. Hint given in feedback Answer:Correct

Explanation

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Question 15

A 0.511 g sample of a nonvolatile, yellow crystalline solid dissolves in 13.0 g of benzene, producing a solution that freezes at 5.39 oC. Find the molar mass of the yellow solid. The following may be useful: The freezing point of benzene is 5.51 oC and the freezing point depression constant, kf, is 4.90 oC/m. Hint given in feedback Answer:Correct

Explanation

Procedure:
(1) Solve ΔTf = mkf for m and find m.
(2) Recall m = mol solute/ kg solvent.
(3) Multiply m by kg solvent to get moles solute (yellow solid).
(4) Recall the definition of molar mass is g/mol. Simply divide the mass of the solute by the number of moles this mass corresponds to.

Question 16

A 4.60% solution by mass of an enzyme in water has an osmotic pressure of 17.9 torr at 286 K. Calculate the molar mass of this enzyme. (Assume that the density of the solution is 1.00 g/mL.) Hint given in feedback Answer:Correct

Explanation

Procedure:
(1) Solve π = MRT for M and find M.
(2) Recall M = mol solute/L solution.
(3) Multiply M by L solution to get moles solute.
(4) Recall the definition of molar mass is g/mol. Simply divide the mass of the solute by the number of moles this mass corresponds to.

Question 17

A saline solution contains 6.1 g of NaCl in 1.00 kg of water. Assuming an ideal value for the Van’t Hoff factor, i, calculate the freezing point of this solution in oC. Answer:Correct

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Question 18

A saline solution contains 7.1 g of NaCl per 0.80 liter of solution. Assuming an ideal value for the van’t Hoff factor, i, calculate the osmotic pressure (in atm) of this solution at 291 K. Answer:Correct

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Question 19

A reverse osmosis unit is used to obtain drinkable water from a source that contains 0.600 g NaCl per liter. What is the minimum pressure (in torr) that must be applied across the semipermeable membrane to obtain water? Assume room temperature or about 298 K. (Don′t forget the van′t Hoff factor!) Answer:Correct

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Question 20

When 0.030 mol of HCl dissolves in 100.0 g of benzene, the solution freezes at 4.04 oC. When 0.030 mol of HCl dissolves in 100.0 g of water, the solution freezes at -1.07 oC. Use data in the following table to calculate the expected molality of HCl in each solvent, ignoring the van’t Hoff factor. Check the 3 statements which are correct. CHEM125 - CHAPTER 12 - REGULAR HOMEWORK 2 Select one or more:a. The calculated molality of HCl in benzene is less than 0.30 m.b. The calculated molality of HCl in benzene equals 0.30 m. Correctc. The calculated molality of HCl in benzene is greater than 0.30 m.d. The results imply that HCl is moderately dissociated in BENZENE.e. The calculated molality of HCl in water equals 0.3 m.f. The calculated molality of HCl in water is between 0.3 m and 0.4 m.g. The calculated molality of HCl in water is between 0.5 m and 0.6 m. Correcth. The calculated molality of HCl in water equals 0.60 m.i. The results imply that HCl is completely dissociated in WATER.j. The results imply that HCl is mostly dissociated in WATER. Correct

Explanation

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Question 21

A mixture contains 25 g of cyclohexane (C6H12) and 44 g of 2-methylpentane (C6H14). The mixture of liquids is at 35 oC . At this temperature, the vapor pressure of pure cyclohexane is 150 torr, and that of pure 2-methylpentane is 313 torr. Assume this is an ideal solution. What is the mole fraction of cyclohexane in the liquid phase? Answer:Correct

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Question 22

What is the vapor pressure (in torr) of cyclohexane above the solution? Answer:Correct

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Question 23

What is the mole fraction of cyclohexane in the vapor phase? Answer:Correct

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Question 24

At a pressure of 760 torr, formic acid (HCOOH; boiling point = 100.7 oC) and water (H2O; boiling point = 100.0 oC) form an azeotropic mixture, boiling at 107.1 oC, that is 77.5% by mass formic acid. At the boiling point of the azeotrope (107.1 oC), the vapor pressure of pure formic acid is 917 torr, and that of pure water 974 torr. If the solution obeyed Raoult’s law for both components, what would be the vapor pressure (in torr) of formic acid at 107.1 oC? Answer:Correct

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Question 25

If the solution obeyed Raoult’s law for both components, what would be the vapor pressure of the water (in torr) at 107.1 oC? Answer:Correct

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Question 26

If the solution obeyed Raoult’s law for both components, what would be the total vapor pressure (in torr) at 107.1 oC? Answer:Correct

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Question 27

Check those statements which are true regarding the above azeotropic mixture. Select one or more:a. The actual total pressure is 760 torr which is smaller than the vapor pressure expected by Raoult’s law. Thus, there is a negative deviation in the vapor pressure. Correctb. For a negative deviation, the attractive forces between water and formic acid must be larger than those in the the pure substances. Correctc. For a negative deviation, the attractive forces between water and formic acid must be smaller than those in the the pure substances.d. The actual total pressure is 760 torr which is greater than the vapor pressure expected by Raoult’s law. Thus, there is a positive deviation in the vapor pressure.e. For a positive deviation, the attractive forces between water and formic acid must be weaker than those in the the pure substances.f. For a positive deviation, the attractive forces between water and formic acid must be stronger than those in the the pure substances.

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