CHM 1046. Chapter 12 Homework Solutions.

Problems: 2, 5, 8, 23, 32, 34, 42, 52, 64, 68, 70, 72, 78, 82, 87, 94

2)A solution is a homogeneous mixture, that is, a mixture that has the same composition throughout.

The solvent is the major component of a solution. The solutes are the minor components of a solution.

5) Entropy is a measure of the amount of disorder in a system. The more disordered the system is, the larger the value for entropy.

In general, a process that increases the amount of disorder in a system is more likely to occur than one that does not. Since forming a solution increases the amount of disorder, entropy favors the formation of a solution.

8) The statement "like dissolves like" means that polar solutes tend to dissolve well in polar solvents, and that nonpolar solutes tend to dissolve well in nonpolar solvents. Polar solutes usually do not dissolve well in nonpolar solvents, and nonpolar solutes usually do not dissolve well in polar solvents.

23)Colligative properties are properties of solutions of nonvolatile solutes with volatile solvents. They are properties that at most depend only on the physical properties of the solvent and the concentration of solute particles.

There are four colligative properties - vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

32)Both CCl4 and CH2Cl2 have a tetrahedral molecular geometry. In both molecules the C - Cl bonds are polar (EN(C) = 2.5, EN(Cl) = 3.0). However, because of the geometry, the contributions of the four polar bonds cancel in CCl4, and the molecule is nonpolar. In CH2Cl2 the contributions of the two polar C - Cl bonds do not cancel, and so the molecule is polar. Polar molecules dissolve better in polar solvents, like water, than nonpolar molecules, so CH2Cl2 should be more soluble in water than CCl4.

34)Molecules a and c (toluene and isobutene) are both hydrocarbons (they contain only carbon and hydrogen atoms). The molecules are nonpolar. Therefore these two substances should dissolve well in hexane (a nonpolar solvent) but not dissolve well in water (a polar solvent). They will interact with the hexane by dispersion forces.

Molecules b and d (sucrose and ethylene glycol) contain several polar - O - H bonds, and so the molecules are polar. Therefore these two molecules should dissolve well in water (a polar solvent) but not dissolve well in hexane (a nonpolar solvent). They will interact with water by hydrogen bonding (a type of dipole-dipole interaction) and dispersion forces.

42)According to Figure 12.11, the solubility of potassium nitrate (KNO3) in water at T = 25. C is about 37 g per 100 g water. So a solution containing 32 g KNO3 per 100 g water is unsaturated.

52)M(KNO3) = 101.1 g/molM(H2O) = 18.02 g/mol

g(KNO3) = 88.4 gmol(KNO3) = 88.4 g 1 mol = 0.8743 mol

101.1 g

g solution = 1500 mL 1.05 g = 1575. g

1 mL

Since g solution = g water + g KNO3,

g water = g solution - g KNO3 = 1575. g - 88.4 g = 1486.6 g

So...

molarity (M) = 0.8743 mol KNO3 = 0.583 mol/L

1.5 L soln

molality(m) = 0.8743 mol KNO3 = 0.588 mol/kg

1.4866 kg water

mass percent KNO3 = 88.4 g KNO3 . 100% = 5.6% KNO3 by mass

1575. g soln

Comments:

1) I have not been that rigorous in significant figures, since I feel Tro has given some of the information in the problem with too few significant figures. On an exam I would be careful to give all of the information to an appropriate number of significant figures.

2) Tro seems not to like to give units for molality - but molality does have units. The units are mol solute/kg solvent (or mol/kg).

3) For dilute aqueous solutions we expect that the numerical value for molarity and molality should be close to one another. This is due to the fact that the density of water is 1.00 g/mL, so 1. kg of water occupies 1. L of volume. In fact, the values here are close, as expected.

4) We have sufficient information in this problem to also calculate the mole fraction of KNO3 in the solution.

64)M(CH3OH) = 32.04 g/molM(H2O) = 18.02 g/mol

g(CH3OH) = 20.2 ml 0.782 g = 15.80 gg(H2O) = 100.0 mL 1.00 g = 100.0 g

1 mL 1 mL

mol(CH3OH) = 15.80 g 1 mol = 0.493 molmol(H2O) = 100.0 g 1 mol = 5.549 mol

32.04 g 18.02 g

a) molarity(M) = 0.493 mol CH3OH = 4.18 mol/L

0.118 L soln

b) molality(m) = 0.493 mol CH3OH = 4.93 mol/kg

0.1000 kg water

c) percent by mass methanol = 15.80 g . 100% = 13.6% methanol

(15.80 + 100.0)g

d) mole fraction methanol = 0.493 = 0.0816

(0.493 + 5.549) mol

e) mole percent methanol = (0.0816) . 100% = 8.16% methanol by moles

68)Let N = naphthalene and H = hexane

MN(C10H8) = 128.2 g/molMH(C6H14) = 86.17 g/mol

Assume 100.0 g of solution. Then there are 10.85 g naphthalene and (100.00 - 10.85) g = 89.15 g hexane.

To find the mole fraction of hexane (the solvent) we need to find the moles of each substance

moles naphthalene = 10.85 g 1 mol = 0.0846 mol

128.2 g

moles hexane = 89.15 g 1 mol = 1.0346 mol

86.17 g

So XH = 1.0346 mol = 0.9244

(1.0346 + 0.0846) mol

Assuming Raoult's law is obeyed, then

pH = XH pH = (0.9244) (151 torr) = 140. torr

70)M(CaCl2) = 111.0 g/molM(H2O) = 18.02 g/mol

Assuming Raoult's law applies, pW = XW pW , or

XW = pW = 81.6 torr = 0.881

pW 92.6 torr

Now, assume that there is a total of 1.000 moles in the system. Then

moles water = 0.881 molg water = 0.881 mol 18.02 g = 15.88 g

1 mol

For the moles of CaCl2 we must account for the fact that this is a strong electrolyte.

CaCl2(s)  Ca2+(aq) + 2 Cl-(aq)

So moles solute particles = (1.000 - 0.881) mol = 0.119 mol

Moles CaCl2 = 0.119 mol particle s 1 mol CaCl2 = 0.0397 mol CaCl2

3 mol particles

g(CaCl2) = 0.0397 mol 110.0 g = 4.37 g

1 mol

Finally, to get the mass percent CaCl2

mass percent CaCl2 = 4.37 g . 100% = 21.6%

(4.37 + 15.88)g

72)Let P = pentane and H = hexane, and assume Raoult's law applies.

ptotal = pP + pH = XP pP + XH pH

We have one equation and two unknowns. But XP + XH = 1, so XP =1 - XH, so

ptotal = (1 - XH) pP + XH pH = pP - XH pP + XH pH

= pP + XH (pH - pP)

Or, finally, XH = (ptotal - pP) = (258 - 425) torr = 0.609

(pH - pP) (151 - 425) torr

And so XP = 1 - 0.609 = 0.391

78)For freezing point depression Tf = Kf m , where m = molality of solute particles

So m = Tf Tf = Tf - Tf = 0.0C - (- 1.3 C) = 1.3 C

Kf Kf (Table 12.8) = 1.86 kg.C/mol

So m = 1.3 C = 0.699 mol/kg

(1.86 kg.C/mol)

There are 150.0 g of solvent (water). The number of moles of solute is

nsolute = 0.1500 kg solvent 0.699 mol = 0.105 mol solute

1 kg

So M = g solute = 35.9 g = 342. g/mol

mol solute 0.105 mol

82)The expression for osmotic pressure is  = [B] RT , where [B] = solute molarity

The moles of hemoglobin is n = 18.75 x 10-3 g = 2.88 x 10-7 mol

6.5 x 104 g/mol

So [B] = 2.88 x 10-7 mol = 1.92 x 10-5 mol/L

0.0150 L

And so  = (1.92 x 10-5 mol/L) (0.08206 L.atm/mol.K) (298.2 K)

= 4.70 x 10-4 atm (760 torr/1 atm) = 0.357 torr

87) = [B] RT , so [B] = /RT

So [B] = 8.3 atm = 0.339 mol/L

(0.08206 L.atm/mol.K) (298.2 K)

The above is the concentration of solute particles. Since the concentration of the ionic solute is 0.100 mol/L, then

i = concentration of solute particles = 0.339 mol/L = 3.39

concentration of solute 0.100 mol/L

94)We first need to find the moles of gas. Assuming the ideal gas law applies

pV = nRT , so n = pV/RTp = 725 torr (1 atm/760 torr) = 0.9539 atm

n = (0.9539 atm) (1.65 L) = 0.06432 mol

(0.08206 L.atm/mol.K) (298.2 K)

Now, from Henry's law, [B] = kH pB , where B = solute

So [B] = (0.112 mol/L,atm) (0.9539 atm) = 0.107 mol/L

Therefore, the volume of solution needed to completely dissolve the solute is

V = 0.06432 mol 1 L = 0.60 L

0.107 mol

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