PROPERTIES OF SOLUTIONS - UNIT 08€¦ · 8. The freezing point of a 1.0m aqueous solution of a substance is approximately -5.6oC. Which of these is the most likely substance dissolved

EXTRA PROBLEMS FOR PRACTICE PROPERTIES OF SOLUTIONS - UNIT 08 ______________________________________________________________

Questions 1 through 27 are multiple choice questions. Questions 28-35 are free response questions and problems.

1. Compared to a 1.0M aqueous solution of glucose, a 1.0M aqueous solution of calcium chloride will have . . .

2. At 298K, 25.00mL of solution containing 27.55mg of protein has an osmotic pressure of 3.22 torr. The molecular mass of the protein, in amu, is . . .

3. What is the relative order of freezing points of these three substances? I. water II. 0.1M aqueous ammonia solution III. 0.1M aqueous ammonium chloride

4. A one thousand liter sample of water contains one gram of iron(III) ions. What is the concentration of Fe3+ (aq) in parts per million?

5. Consider a 0.50M solution of each of the following salts. Which will have the lowest freezing point?

6. What is the vapor pressure of a solution containing 0.500 mol of glucose and 5.00mol of water at 29oC.

7. For a solution containing a nonvolatile solute dissolved in a volatile solvent, what is true of the vapor pressure, boiling point, and freezing point of the solution compared to the pure solvent?

8. The freezing point of a 1.0m aqueous solution of a substance is approximately -5.6oC. Which of these is the most likely substance dissolved in the solution? The molal freezing point constant for water is -1.86 oC/m.

9. What is the mole fraction of water in a solution that contains 32g of methanol in 36g of water?

(a) the same freezing and boiling points. (b) a lower freezing point and a lower boiling point.

(c) a lower freezing point and higher boiling point. (d) a higher freezing point & a higher boiling point.

(e) a higher freezing point and a lower boiling point.

(a) 2,340,000. (b) 159,000. (c) 6360. (d) 254.

(a) III < II < I (b) II < III < I (c) II = III < I

(d) II < III = I (e) I < II < III

(a) 0.001 (b) 0.01 (c) 0.1 (d) 1 (e) 10

(a) NaCl (b) MgCl2 (c) K2SO4 (d) Cr(NO3)3 (e) CaSO4

(a) 30.0 torr (b) 27.3 torr (c) 25.0 torr (d) 2.7 torr

(a) VP = increases BP = increases FP = increases (b) VP = decreases BP = decreases FP = decreases

(c) VP = increases BP = decreases FP = increases (d) VP = decreases BP = decreases FP = increases

(e) VP = decreases BP = increases FP = decreases

(a) CH3CH2OH (b) NaCl (c) Ca(NO3)2 (d) C6H6 (e) Al2(SO4)3

(a) 0.33 (b) 0.50 (c) 0.67 (d) 1.0 (e) 1.1

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10. Which pairs of substances will dissolve in each other? I. CH3OH II. C6H6 III. CH3CH3

11. Adding 1mol of NaCl to 1kg of water lowers the vapor pressure of water more than adding 1mol of C6H12O6. Explain. (a) NaCl is highly polar whereas C6H12O6 is weakly polar and this causes the difference. (b) NaCl has a smaller molar mass than that of C6H12O6 and this causes the difference. (c) NaCl is polar whereas C6H12O6 is nonpolar and this causes the difference. (d) The total solute concentration of NaCl is twice that of C6H12O6 because NaCl forms Na+ and Cl− ions.

12. Which substances will not form a gas upon mixing with an aqueous acid?

13. How many milliliters of 0.40M FeBr3 solution would be necessary to precipitate all of the Ag+ from 30.mL of a 0.40M AgNO3 solution according to the following equation? FeBr3 (aq) + 3 AgNO3 (aq) → Fe(NO3)3 (aq) + 3 AgBr (s)

14. The aqueous solutions of which salt will have the strongest cation-dipole interactions?

15. How does the magnitude of ΔHmix compare with the magnitude of ΔHsolute + ΔHsolvent for exothermic solution processes?

Use the structures of cyclohexane, glucose, pentane, and acetone to answer questions 16 - 18.

(a) I and II only (b) II and III only (c) I & II, I & III, II & III

(d) I and III only (e) I & II, II & III only

(a) NaHCO3 (s) (b) Mg (s) (c) Ca (s) (d) K2CO3 (s) (e) Al2O3 (s)

(a) 10. (b) 20. (c) 30. (d) 40. (e) 50.

(a) NaCl (b) MgCl2 (c) K2SO4 (d) Cr(NO3)3 (e) CaSO4

(a) larger (b) almost the same (c) smaller

cyclohexane

C

C

C

C

C

C

H

H

H

H

H

H

H

H

HH

H

H

acetone

glucose

pentane



16. Pentane is most soluble in . . .

17. When glucose dissolves in water, the strongest intermolecular forces in the solution are . . .

18. Which of the following molecules will not dissolve in water?

Questions 19 and 20 refer to the following statement: When solid ammonium chloride dissolves in water, the solution becomes cool.

19. For such a process, which statement best describes the reason why dissolving occurs spontaneously?

20. Which double-sided arrow in the

following diagram represents the change in enthalpy of solution, ∆Hsolution? The length of the double-sided arrows are proportional to enthalpy changes involved when a solid dissolves to form a solution. (a) A (b) B (c) C (d) D

21. A laboratory procedure requires 0.270 mole of methanol, CH3OH to be added as a solvent. What volume of 1.50M CH3OH is needed to attain this number of moles?

22. Which substance is least soluble in water?

(a) glucose only. (b) glucose and acetone. (c) glucose, acetone, & cyclohexane.

(d) cyclohexane only. (e) cyclohexane and glucose.

(a) dispersion. (b) ion-dipole. (c) hydrogen bonds. (d) ionic. (e) covalent.

(a) acetone (b) cyclohexane (c) pentane

(d) acetone and cyclohexane (e) cyclohexane and pentane

(a) Enthalpy decreases. (b) Enthalpy increases. (c) Vapor pressure decreases. (d) Entropy increases.

(a) 5.55 mL (b) 45.0 mL (c) 405 mL (d) 180 mL

(a) methanol

#

(d) methyl bromide

#

(c) methane

(b) methyl amine

#



23. The reason for the answer to question 22 is . . . (a) methanol forms the strongest hydrogen bonding with the water molecules. (b) methyl amine forms the strongest hydrogen bonding with the water molecules. (c) methane exhibits London dispersion forces with water molecules. (d) methyl bromide has the highest molar mass.

24. How do the slopes of the lines on the solubility graph to the right vary with molecular mass of the gas? Explain the trend. (a) The slopes increase linearly as the molecular weights increase. Increasing

molecular mass leads to greater polarizability of the gas molecules and greater intermolecular attractions between gas and water molecules.

(b) The slopes exponentially increase as the molecular weights increase. Increasing molecular mass leads to greater polarizability of the gas molecules and greater intermolecular attractions between gas and water molecules.

(c) The slopes increase linearly as the molecular weights decrease. Decreasing molecular mass leads to greater polarizability of the gas molecules and greater intermolecular attractions between gas and water molecules.

(d) The slopes exponentially decrease as the molecular weights decrease. Decreasing molecular mass leads to greater polarizability of the gas molecules and greater intermolecular attractions between gas and water molecules.

25. A pressure of 3.5 atm in CO2 is required to maintain a 0.12M CO2 concentration in soda. Calculate the Henry’s law constant for CO2.

26. On a clear day at sea level, the partial pressure of N2 isn air is 0.78 atm at 25oC. Under these conditions, the concentration of N2

in water is 5.3x10−4M. What is the partial pressure of N2 when the concentration in water is 1.1x10−3M?

27. Consider two solutions: (1) 10 g of glucose (C6H12O6) in 1L of water and (2) 10 g of sucrose (C12H22O11) in 1L of water. Which of the following is true? (a) The glucose solution has the higher vapor pressure. (b) The sucrose solution has the higher vapor pressure. (c) Both solutions have the same vapor pressure. (d) There is not enough information to compare vapor pressures.

28. 175g of calcium chloride is dissolved in 975g water. The density of the solution is 1.10 g/mL. For water at 25oC: VP = 23.76 torr FP constant = 1.86oC/m BP constant = 0.51oC/m a. What is the vapor pressure of the solution at 25oC?

Moles CaCl2 = 1.58 moles Moles H2O = 54.2moles CaCl2 dissociates into 3 particles so 3 x 1.58 = 4.74 moles dissolved particles Psoln = solvent mole fraction x Psolvent

Psoln = (54.2 moles ÷ (54.2 + 4.74)moles) x 23.76 torr = 21.9 torr b. At what temperature will the solution freeze?

ΔT = 3 (1.86oC/m) (1.58moles ÷ 0.975kg) = 9.01oC FP = -9.04oC c. At what temperature will the solution boil?

ΔT = 3 (0.51oC/m) (1.58moles ÷ 0.975kg) = 2.48oC BP = 102.48oC d. What will be the osmotic pressure?

Total mass = 975g + 175g = 1150g 1150g x 1ml/1.10g x 1L/1000mL = 1.05L P = 3 (1.58mol ÷ 1.1.05L) (0.0821 L-atm/mol-K) (298K)

(a) 29.2 (b) 0.42 (c) 2.9x101 (d) 3.4x10−2 (e) 4.2

(a) 0.63atm (b) 0.78 atm (c) 1.0 atm (d) 2.1 atm (e) 1.6 atm



29. An unknown compound contains only C, H, and O. Combustion analysis of the compound gives mass percents of 31.57%C and 5.30%H. A solution made by dissolving 10.56g of the compound in 25.0mL of water freezes at -5.20oC. Determine the empirical formula, molar mass, and molecular formula of the unknown compound. Assume the compound is a nonelectrolyte. The freezing point constant for water is 1.86oC/m.

Empirical formula 31.57g C = 2.63 mol ÷ 2.63 = 1 5.30g H = 5.30 mol ÷ 2.63 = 2 63.13g O = 3.95 mol ÷ 2.63 = 1.5 C2H4O3

5.20 oC = (1.86oC/m) m m = 2.80 m 0.0250kg x 2.80mol/kg = 0.0700mol 10.56g / 0.0700mol = 151 g/mol (C2H4O3) x = 151 g/mol x = 2 C4H8O6

30. Answer the following questions about these laboratory observations. Solid ammonium chloride dissolves in water with a marked decrease in temperature. Calcium chloride solid dissolves in water with a marked increase in temperature. Little or no temperature change is observed when solid sodium chloride dissolves in water. a. Write an equation that describes the dissolving process of ammonium chloride. b. Is the dissolving of calcium chloride endothermic or exothermic?

Exothermic c. Describe the opposing forces of attraction that are at work in the dissolution of calcium chloride. Which are greater? Why?

Ionic bonds are broken and ion-dipole forces are formed. More energy to form than to break so EXO

d. What can be said about the opposing forces of attraction when sodium chloride dissolves in water? Why? ion-dipole energy about equal to ionic bonds breaking

e. Use the observation for ammonium chloride to discuss these seemingly contradictory statements: (i) Thermodynamically, exothermic processes tend to be spontaneous. (ii) Most processes occur spontaneously when there is an increase in entropy.

Want least energy so exothermic from an energy standpoint is most stable. Want most possible energy states so increase in entropy is favorable.

31. The molecular formula of an unknown compound is determined by combustion analysis and freezing point depression. A solution containing 0.496g benzoic acid (C6H5COOH), and 25.0g camphor (C10H16O), freezes at 173.3oC. The freezing point of pure camphor is 179.8oC. An unknown molecular compound is found to contain 80.77%C, 3.846%H, and 15.38%O. A solution of 0.243g of unknown and 15.1g of camphor melts at 176.7oC. a. What is the empirical formula of the unknown compound?

C7H4O b. What is the freezing point constant for camphor? Specify units.

m = 0.163m ΔT = (176.7 - 173.3oC) = Kf (0.163m) Kf = 39.9oC/m

c. What is the concentration of the unknown compound in camphor in units of molality? (179.8 - 176.7) = (39.9oC/m) m m = 0.0777

d. What is the molar mass of the unknown compound? mole unk = (0.0777 mol unk / kg camp) 0.0151 kg camp = 0.00117 mol unk 0.243 g / 0.00117 mol = 208 g/mol

e. What is the empirical formula of the unknown compound? 104 x 2 = 208 so molec = C14H8O2



32. Three unknown metals are labeled A, B, and C. One is silver, one is aluminum, and the other is nickel. Using only aqueous solutions of 0.50M iron(II) nitrate and 1.0M hydrochloric acid, write a short, concise, experimental procedure, the results of which will be sufficient to identify each of the unknown metals. Tell whether you would expect to see and what the observation means. Write net ionic equations to illustrate your answers.

Add one drop of acid to each metal and observe the results. Silver is the only metal of the three that will NOT react. The other two metals will react to produce bubbles of hydrogen gas. 2 Al(s) + 6 H+ → Al3+ + 3 H2(g) Ni(s) + 2 H+ → Ni2+ + H2(g) Because both aluminum and nickel are above hydrogen in the activity series, both metals will react with H+(aq). Silver will not react with aqueous acid because it is below hydrogen on the activity series.

Add a drop of iron(II) nitrate to freshly cleaned pieces of the other two metals. The nickel will not react but the aluminum will oxidize and the iron will reduce. 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s)

Because aluminum is above iron on the activity series, aluminum metal will react with aqueous iron(II) ions. Nickel will not react with iron(II) nitrate because nickel is below iron on the activity series.

33. Two important vitamins are shown in the figure below.

Tell whether each vitamin is more likely to be water soluble or fat soluble. Justify your reasoning.

Vitamin C, with its many OH groups, will be soluble in water because it can readily form H-bonds. Vitamin A will be soluble in fats because the molecule is nonpolar.



a. What is the principle intermolecular force acting when a nonpolar substance dissolves in a nonpolar solvent?

D i spe r s i on f o r ce s a re t he principle intermolecular force acting when a nonpolar solute dissolves in a nonpolar solvent.

b. Describe the principle intermolecular force(s) acting when a water-soluble vitamin dissolves in water?

Hydrogen bonds c. Identify one aspect that the space-filling models in the

figure represent accurately and one aspect that limits their usefulness.

T h e s p a c e - f i l l i n g m o d e l represents the three-dimensional shape of the molecule very well. It does not show polarity of the molecule of the nature of the delocalized electrons.

d. Identify one aspect that the structural formulas represents accurately and one aspect that limits their usefulness.

Structural formulas give a clear picture of which atoms are present as well as atom connectivity. They do not depict the molecular geometry or their electronic nature.

34. The figure shows the dynamic equilibrium in a saturated solution with excess ionic solute. a. Describe the forces that must be overcome for the salt to dissolve.

The ionic bonds that bind the solid crystal and the hydrogen bonds that hold the water molecules together must be broken.

b. Are the larger ions depicted most likely cations or anions? Give two reasons for your answers.

The larger ions are anions. Anions tend to be larger than cations because of electron-electron repulsion of the extra electrons in the valence shell. Also, the larger ions are shown attracted to the positive dipoles of the water molecule.

c. Describe the principle force that holds the ions in solution.

Ion-dipole forces, the attraction of the ions to the dipoles of water molecules, hold the particles in solution.

d. Under what conditions will this system never reach equilibrium? Explain.

The system will never reach equilibrium if the dissolving process is exothermic. Entropy increases as an ionic substance dissolves because the very ordered crystal lattice of the ionic solid becomes dispersed throughout the liquid. Increasing entropy drives change. The tendency of a system to move toward lower potential energy also drives change. If the system is exothermic, the driving forces of increased entropy and decreased potential energy work together, so equilibrium will never be reached.



e. Comparing a salt solution to pure water with both systems containing the same amount of water, which system will require more heat to completely evaporate the system to dryness? Explain.

A salt solution requires more heat to evaporate than does pure water because the extra heat is required to break the relatively strong solute-solvent interactions, in this case the ion-dipole forces.

35. When HNO2 is dissolved in water, it partially ionizes according to the equation: HNO2 ⇌ H+ + NO2−. A solution is prepared

that contains 7.050g HNO2 in 1.000kg of water. Its freezing point is −0.2929oC. Calculate the fraction of HNO2 that has ionized.

7.050g HNO2 x 1mol HNO2/47.02g HNO2 = 0.1499mol HNO2 0.1499mol HNO2 ÷ 1.000kg = 0.1499m HNO2 ΔT = iact kf m 0.2929oC = iact (1.86oC/m) (0.1499m) iactual = 1.050 fraction ionized = iactual − 1 = 1.050 − 1 = 0.050


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PROPERTIES OF SOLUTIONS - UNIT 08€¦ · 8. The freezing point of a 1.0m aqueous solution of a substance is approximately -5.6oC. Which of these is the most likely substance dissolved