Properties of Solutions

[Pages:5]Chem 116 POGIL Worksheet - Week 4 Properties of Solutions

Key Questions

1. Identify the principal type of solute-solvent interaction that is responsible for forming the following solutions: (a) KNO3 in water; (b) Br2 in benzene, C6H6; (c) glycerol, CH2(OH)CH(OH)CH2OH, in water; (d) HCl in acetonitrile, CH3CN [HCl does not form ions in CH3CN].

(a) ion-dipole (b) London dispersion (c) hydrogen bonding (d) dipole-dipole

2. For the following carboxylic acids, predict whether solubility will be greater in water or carbon tetrachloride, and give your reasoning: (a) acetic acid, CH3CO2H, (b) stearic acid, CH3(CH2)16CO2H.

(a) Acetic acid's -OH groups make hydrogen bonding possible, which is compatible with solvent water. Carbon tetrachloride has only London dispersion forces, which are less compatible. Therefore, acetic acid is more soluble in water than carbon tetrachloride.

(b) Stearic acid has a very long chain and much higher London dispersion forces than acetic acid. This makes it more compatible with carbon tetrachloride, despite the potential for hydrogen bonding (which is largely mitigated by the long chain getting in the way). Therefore, stearic acid is more soluble in carbon tetrachloride than water.

3. Hexane (C6H14) and heptane (C7H16) are miscible in all proportions with )Hsoln .0. (a) Why are these two liquids miscible with each other?

They have very similar London dispersion forces, owing to their similar molar masses.

(b) Why is )Hsoln .0 for this pair of liquids?

The intermolecular forces of attraction in the neat liquids are so similar to each other that little change occurs on mixing. It is the change in intermolecular attraction strength that is principally responsible for the sense and magnitude of )Hsoln.

(b) Why do they spontaneously form solutions, given that )Hsoln .0?

Mixing is a more disordered state than exists in the separate neat liquids. The increase in entropy is the driving factor in making solution formation spontaneous in this case.

4. The solubility of N2 at p(N2) = 1 atm is 1.75 x 10-3 g/100 mL of water. What is the solubility in water at an air pressure of 2.51 atm, the pressure at 50 ft below the surface of the water? Air is 78.1 vol-% N2. [Hint: What is the partial pressure of N2(g) when the air pressure is 2.51 atm?]

From Dalton's Law of Partial Pressures, at air pressure of 2.51 atm, the partial pressure of N2 is

= (0.781)(2.51 atm) = 1.96 atm From Henry's Law, the solubility is

5. Calculate the molality of ethanol, C2H5OH (m.w. = 46.06) in a solution prepared by dissolving 5.00 g of ethanol in 25.00 g of water.

6. Calculate the total molality of all ions in a solution prepared by dissolving 20.0 g of (NH4)2SO4 in 95.0 g of water. [f.w. (NH4)2SO4 = 132 u] (NH4)2SO4(s) 6 2NH4+(aq) + SO42?(aq)

7. Consider a 2.00 m solution of sugar in water at 25.00 oC. (a) What is the value of the mole fraction of water in this solution? [Hint: Imagine that the solution was made up with exactly 1 Kg of water.] (m.w. H2O = 18.02 u) To calculate mole fraction, we need the numbers of moles of water and of sugar. If we assume that exactly 1000 g of water were used, then we already know that the solution contains 2.00 moles of sugar. All we need is the number of moles in 1000 g of water, and then we can calculate P(H2O).

mol sugar = 2.00 mol sugar

mol H2O = (1000 g H2O)(1 mole H2O/18.02 g H2O) = 55.49 mol H2O P(H2O) = 55.49 mol/(55.49 + 2.00) mol = 55.49/57.49 = 0.9652

(b) Calculate the vapor pressure above a 2.00 m solution of sugar in water at 25.00 oC, given that the vapor pressure of pure water at this temperature is 23.76 mm Hg. Psoln = P(H2O) ? P o(H2O) = 0.9652 ? 23.76 mm Hg = 22.93 mm Hg

8. Calculate the expected vapor pressure above a 2.00 m solution of Na2SO4 in water at 25.00 oC. Compare this result to what you found in part a.

We must take account of the dissociation of Na2SO4(s): Na2SO4(s) 6 2Na+(aq) + SO42?(aq)

Again, assume a solution prepared with exactly 1000 g of water.

mol ions = 3 ? mol Na2SO4 = 3 ? 2.00 mol = 6.00 mol P(H2O) = 55.49 mol/(55.49 + 6.00) mol = 55.49/61.49 = 0.9024 Psoln = P(H2O) ? P o(H2O) = 0.9024 ? 23.76 mm Hg = 21.44 mm Hg The amount by which the vapor pressure of water has been lowered is almost three times greater in the Na2SO4 solution, compared to the sugar solution. 9. What are the partial pressures and total vapor pressure above a solution at 20.0 oC made by mixing 12.5 g benzene (C6H6) with 44.2 g toluene (C6H5CH3). At 20.0 oC, P o(C6H6) = 74.7 torr and P o (C6H5CH3) = 22.3 torr. [m.w. C6H6 = 78.11; m.w. C6H5CH3 = 92.14] mol C6H6 = (12.5 g C6H6)(mol C6H6/78.11 g C6H6) = 0.160 mol C6H6 mol C6H5CH3 = (44.2 g C6H5CH3)(mol C6H5CH3/92.14 g C6H5CH3) = 0.480 mol C6H5CH3 total moles = 0.160 mol + 0.480 mol = 0.640 mol

P(C6H6, soln) = 0.160 mol/0.640 mol = 0.250 P(C6H5CH3, soln) = 1 ? 0.250 = 0.750 P(C6H6) = (0.250)(74.5 torr) = 18.7 torr P(C6H5CH3) = (0.750)(22.3 torr) = 16.7 torr

Pt = (18.7 + 16.7) torr = 35.4 torr

10. In terms of mole fractions, what is the composition of the vapor in the previously described benzene-toluene mixture?

We previously found

P(C6H6) = (0.250)(74.5 tor) = 18.7 torr P(C6H5CH3) = (0.750)(22.3 torr) = 16.7 torr Pt = (18.7 + 16.7) torr = 35.4 torr Using these values, we can calculate the mole fractions in the vapor as follows:

P(C6H6, vap) = 18.7 torr/35.4 torr = 0.528

P(C6H5CH3, vap) = 16.7 torr/35.4 torr = 1 ? 0.528 = 0.473

Notice that more volatile benzene has an increased mole fraction in the vapor.

11. Pure benzene has a freezing point of 5.5 oC and a boiling point of 80.1 oC. What are the

expected freezing point and boiling point for a 0.15 m solution of a nonvolatile solute in benzene? For benzene, Kf = 5.12 oC/m and Kb = 2.53 oC/m.

)Tf = (5.12 oC/m)(0.15 m) = 0.77 oC

T soln f

=

T f

? )Tf

= (5.5 ? 0.77) oC = 4.7 oC

)Tb = (2.53 oC/m)(0.15 m) = 0.38 oC

T soln b

=

T b

+

)Tb

=

(80.1 + 0.38) oC = 80.5 oC

12. When 45.0 g of an unknown nonvolatile nonelectrolyte is dissolved in 500.0 g of water, the resulting solution freezes at -0.930 oC. What is the molar mass of the unknown substance? Kf = 1.86 oC/m for water.

)Tf = 0.930 oC m = )Tf/Kf = 0.930 oC/1.86 oC/m = 0.500 m = 0.500 mol X/kg H2O

13. What is the osmotic pressure of a 0.100 M glucose solution in torr at 25.0 oC?

B = MRT = (0.100 mol/L)(0.0821 LAatm/KAmol)(298 K) = 2.45 atm ? (760 torr@atm-1) = 1860 torr

14. Sea water is approximately 0.60 M NaCl. What is the minimum applied pressure that must be exceeded to achieve water purification by reverse osmosis at 25 oC?

We must use the molarity of ions, not the stated molarity of NaCl (analytical concentration of NaCl).

NaCl(s) 6 Na+(aq) + Cl?(aq)

molarity of ions = 2 ? CNaCl = 2 ? 0.60 M = 1.20 M

B = MRT = (1.20 mol/L)(0.0821 LAatm/KAmol)(298 K) = 29.36 atm = 29.4 atm

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