CHEMISTRY 113 HOMEWORK 4 NAME___________________________

Due Wednesday Feb 7, 2001 Computer Homework: Solutions: all

1. Pentane, C5H12, and heptane, C7H16, are two hydrocarbon liquids present in gasoline. At 20oC, the vapor pressure of pentane is 420 torr and the vapor pressure of heptane is 36.0 torr.

  1. What is the vapor pressure at 20oC of a solution prepared by mixing equal masses of the two liquids? MW pentane = 72.15, MW heptane = 100.20 ; assume 100 g of each.

Pt = XpPpo +Pho

= (100 g/ 72.15 g/mol) / (1.386 mol + 100 g/100.20 g/mol) [420 torr]

+ 0.998 mol/(1.386 + 0.998 mol) [36.0 torr]

= (1.386 / 2.384)[420 torr] + (0.998 / 2.384)[ 36.0 torr] = 0.5814(420) + 0.4186(36.0 =

= 244.18 torr + 15.07 torr = 259.251 torr

PT = |_259 torr______

b. What is the mole percent composition of the vapor phase at 20oC?

n proportional to P so

Xp = 244.18/ 259.25 = 0.9418

% pentane = |_94.2%__

Xh = 15.07 / 259.25 = 0.0581

%heptane = |_5.81 %__

Note how the vapor phase is enriched in the more volatile chemical.

2. A solution prepared by dissolving 9.2 g of a nondissociating solute in 1.000 mole of chloroform, CHCl3, has a vapor pressure of 511.0 torr at 25oC. The vapor pressure of pure chloroform at the same temperature is 526.0 torr. Calculate

a) the mole fraction of the solute,

Pc = XcPco = 511.0 torr = Xc(526.0 torr); Xc = 511.0 / 526.0 = 0.97148

Xs = 1.000 – 0.97148 = 0.028517

mol. frac. = |_0.02852__

Note that you must get the mol fraction of the solvent first because it is responsible for the vapor pressure

b) the number of moles of solute, and

XC = 1.000 / (n + 1.000 ) = 0.97148;

1.000 = 0.97148n + 0.97148

0.97148n = 1.000 – 0.97148 = 0.02852

n = 0.029354 mol

mol = |_0.02935 mol __

c) the molecular weight of the solute.

n = mass/MW; MW = mass/n = 9.2 g / 0.02935 mol = 313.4 g/mol

MW = |_3.1 x 102 g/mol___

3. Glycerol, C3H5(OH)3, MM= 92, is a nonvolatile liquid very soluble in water. A solution is made by dissolving 46.0 g of glycerol in 250 g of water. Calculate the freezing point and boiling point of this solution

DTf = Kfm = 1.86 deg kg/mol x (46.0 g/92 g/mol) / 0.250 kg = 1.86 x 2.0 = 3.72o

0.0o – 3.72o = - 3.7o

F.P. = _- 3.7o C____

DTb = Kbm = 0.51 deg kg/mol x (46.0 g/92 g/mol) / 0.250 kg = 0.51 x 2.0 = 1.02o

100.0o + 1.02o = 101.0o C

B.P. = 101.0o C___

4. A 0.520 m aqueous solution of KCl is 83.0 % dissociated. (100 – 83) = 17 à 83 + 83

a) What is the freezing point of the solution?

DTf = iKfm = (183.0%/100%)1.86 deg kg/mol x (0.530 m) = 1.770o

0.00o – 1.77 = - 1.77o

F.P. = _-1.77o C____

b) Approximate the osmotic pressure of the solution.

Assume that m ~ M; also the T = 25oC

P = iMRT = 1.83 x 0.520 M x 0.08206 Latm/mol deg x 298 K = 23.27 atm

p = _23.3 atm____

 

5. Calculate the molecular weight of a polypeptide if 0.400 g in 1.00 L aqueous solution has an osmotic pressure of 3.74 torr at 27 oC.

P = MRT = mass/(MW x V) x RT;

MW = (mass x RT)/ (P x V)

= 0.400g x 0.08206 (Latm/mol deg x 300 K) / [(3.74 torr/760 torr/atm) x 1.00L]

= 9.8472 / 0.00492 = 2001 g/mol

MW = |_2.00 x 103 g/mol_

 

6. The osmotic pressure of a saturated solution of codeine (MW = 317) in water is 0.641 atmospheres at 25oC. Calculate the molarity of a saturated solution of codeine.

P = MRT; M = P / RT= 0.641 atm/ (0.08206 Latm/moldeg x 298 K) = 0.0262 M

 

M = |_0.0262 M____

7. An aqueous solution of a weak electrolyte HX, with a concentration of 0.125 m has a freezing point of –0.261 oC. What is the percent ionization of the compound?

DTf = iKfm; i = DTf / Kfm = 0.261o / (1.86 deg kg/mol x 0.125 mol/kg) = 1.123

HX = H+ + X-; (1.00-x) à x + x; i = 1.123 = 1.00 = x; x = 0.123

0.123 x 100% = 12.3% dissociated

% = |_12.3%___