Homework Answers Kotz & Treichel 1997

The division line is missing in equations below for some strange computer reason.

14.13
2.65 g malic acid dissolved in 0.500L, 500 g, of water. What is the molarity, molality, mole fraction and weight percent of malic acid in the solution.

ans. The molecular weight of malic acid is 134.1 g/mole.

Concentration moles/liter:
[malic acid] =

molality:

mole fraction:
moles water = ,

moles malic acid = ;

total moles = 0.019 + 27.76 = 27.78 moles.

mole fraction malic acid =

mole fraction water =

14.28
moles/0.150L = 0.1. moles = 0.015. Each sodium sulfate gives three ions so we need 0.015/3 = 0.005 moles sodium sulfate. The molecular weight of sodium sulfate = 142 g/mole so the weight needed is 142*0.005 = 0.071 g = 71 mg.

14.29
Lets assume that sea water is essentially pure water.
moles Li^+ in one liter of pure water = 0.18 ppm * 1000g / 6.94 g per mole = 0.0000259 moles per kilogram water, or 0.0000259 molal.

14.31
Thinking of heat as a reagent then (c) "Raise the temperature and add more NaCl." is the answer.

14.32
Given the solubility of potassium nitrate as a function of temperature (see graph) what is the sign of the enthalpy of solution?

Ans. We can treat heat as a reagent and use LeChatlier's principle. As the temperature is raised more goes into solution so if we write adding more heat would drive the reaction to the right. Thus, the enthalpy of solution, , would be positive.

14.35
Obviously, water, being the most polar solvent, should have the most negative enthalpy of solvation.

14.36

The problem here is that I have two conflicting sets of enthalpies for the solvation reaction and I cannot tell which is correct, or why such a state of affairs should occur! I put one table of enthalpies below and it should be correct, but it dissagrees with what you might calculate from the tables in the back of the K&T text. However, the trends with ion size are the same.

There are two competing effects:

(1) Larger ions have to carve out a larger volume in water breaking relatively more H-bonds and requiring some "energy".

That is, larger ions displace more waters and break more H-bonds. This requires enthalpy into the system so > O. Smaller ions form a stronger H-bond system.

(2) Surrounding each ion are H-bonding structures lowering the energy and releasing heat so < O.

Which of these effects wins out is hard to say. We may use the thermo chemical data below to support the view that the second, formation of solvent shells of water, is the more important.

14.37
The partial pressure in lungs is approximately 40 torr (mm Hg). How much O2(g) can disolve in water under this partial pressure?

molality of oxygen = Kh*P(O_2) = 1.66 x 10^-6 M/torr * 40 torr = 6.65 x 10^-5 molal and since this is water where 1 kg = 1 liter = 6.65 x 10^-5 molar.

14.38
Kh(O2) = 1.66 x 10^-6 M/mmHg at 298K. As temperature increases Kh decreases since the solubility of oxygen in water decreases. At 50C (a) is the best guess.

14.39
Another plugin problem. Sg = K_H P(CO_2) = 0.0506M = 4.48 x 10^-5 M/torr x P(CO_2). P(CO_2) = (0.0506M) / (4.48 x 10^-5 M/torr) = 1130 torr (mmHg) = 1130 mmHg/760 mmHg/atm = 1.49 atm.

14.41
Assuming ideal behavior Raoult's Law is Pg = Xg P^o. Ethlyene glycol is assumed to have no vapor pressure and the vapor pressure of water varies with temperature being 35.7 mmHg at 32 C.
The number of moles of water = 500/18.02 = 27.76 moles.
The number of moles of glycol = 35.0 g/62.06 = 0.564 moles.
The mole fraction of water = 27.76/(27.76 + 0.564) = 0.980.
The vapor pressure of water from the mixture = 0.980*35.7 mmHg = 35.0 mmHg (torr).

14.45
What is the molar mass of 10 g of a non-volatile solute dissolved in 100 g of benzene if the vapor pressure of pure benzene is 121.8 torr at 30C and the vapor pressure above the mixture is 113.0 torr?

P(benzene) = 113 mmHg = X(benzene)*121.8 mmHg; Mole fraction of benzene = X(benzene) = 113/121.8 = 0.928. The molecular weight of benzene = 78.11 g/mole.
X(benzene) = 0.928 = (moles benzene)/(moles benzene + moles solute) = 1.28/(1.28 + moles solute). Solving for moles solute gives 0.0993 moles. Mw = 10/0.0993 = 100.7 g/mole.

14.51
The boiling point elevation for a solution of glycerol in water is 4.3 C. = +0.5121*m(molal). m(molal) = 8.4 moles/kg. Moles solute = 8.4 m/kg*0.75 kg = 6.3 moles. Wt glycerol = Mw*6.3 = 92.1 g/mole * 6.3 mole = 580 g
moles water = 750g/18.02 g/mole = 41.62 moles. Total moles = 41.62 + 6.3 = 47.9.
mole fraction glycerol = 6.3/47.9 = 0.13.

14.54
Arrange the aqueous solutions in order of increasing boiling point.

Increasing bp (a), (d), (c), (b) where the number of dissolved species has been taken into account.

14.56
0.255 g of C_10H_8Fe added to 11.12 g benzene. Mw(C_10H_8Fe) = 184.01 g/mole and Mw(benzene) = 78.11 g/mole. The = 80.26 - 80.10 = 0.16 C = +2.53*m(molal). m(molal) = 0.16/2.53 = 0.063. molality = moles C_10H_8Fe/kg benzene and so moles C_10H_8Fe = 0.063*0.01112 = 0.00070 moles. The molecular weight = wt/moles = 0.255 g/0.00070 moles = 363 g/mol. Since the empirical formula give a molecular weight of 184 and 363.184 = 1.97 the molecule must be a dimer that dissociates into two molecules in solution.

14.60
(a) Freezing point of glycol-water mixture = -15C = -1.86* m(glycol). m(glycol) = 8.06 molal = moles glycol/kg water. moles glycol = 5kg * 8.06 = 40.3 moles. Mw(glycol) = 62.07 g/mole so the wt glycol = 40.4 moles * 62.07 g/mole = 2,500 g = 2.5 kg. The density of glycol is 1.11 g/mL so this is about 2.25 Liters.

(b) The boiling point will be increased according to = +0.5121*m = 0.5121*8.05 = 4.15C. The bp of water containing this glycol = 104.2 C. Hence the idea of using gylcol in the summer to keep the car from "getting too hot" and boiling over.

14.66
0.10 g naphtalene,, is added to 10.0 g biphenyl,. The f.p. of biphenyl is 70.03C and the freezing point depression constant for biphenyl is Kfp = -8.00 C/molal.
The freezing point of the mixture is found to be 69.40C. What is the molar mass of naphtalene?

= 69.40 - 70.03 = -8.00*m
m = 0.079 = moles nap/kg biphenyl = (0.1/Mw)/0.010kg.
Solve for Mw(nap) = 0.1g/0.079 moles/kg*0.010 kg = 127 g/mole. (The actual Mw = 128.16.)

14.70
There are effectively 1.9 ions in the solution whose concentration for NaCl is c = 1.9*0.16 moles/L, The gas constant is 0.082 L-atm/mole-K and the temperature 310K (37C). = 1.9*0.16*0.082*310 = 7.7 atm.

14.71
The osmotic pressure of 1.0 g bovine insulin in a liter of water is measured to be 3.1 mmHg = 0.00408 atm at 298K. c = /R*T = 0.00408/0.082*298 = 0.00017 moles/liter. This is due to 1.0 g insulin in 1L so the Mw = (1.0 g/L)/0.00017 moles/L = 5990 g/mole.

14.105
A solution of 5.0g acetic acid in 100g benzene freezes at 3.37C. The Kfp = -8.00 for benzene.
A solution of 5.0g acetic acid in 100g water freezes at -1.49C. The Kfp = -1.86 for water.
Using this data calculate the molecular weights of acetic acid in the two experiments.

In benzene
Pure benzene freezes at 5.50C
= 3.37 - 5.50 = Kfp*m = -5.12*(5g/Mw)/0.100L. Solve for Mw = 5.12*5/(2.13*0.1) = 120 g/mole

In water
Pure water freezes at 0.0C
= -1.49 - 0.0 = Kfp*m = -1.86*(5g/Mw)/0.100L. Solve for Mw = 1.86*5/(1.49*0.1) = 62 g/mole

The thing that makes most sense is that in benzene a dimer of acetic acid is formed as in . The arguement that this occurs would be that in water there are many waters surrounding (the solvation sphere) acetic acid stablizing it as single solvated species. In benzene, a non-polar solvent, no H-bonding can occur between benzene and acetic acid, but acetic acid molecules would be attracted and form a lower energy state by H-bonding. The simplest H-bonding species would be the dimer.