Chemistry 1A Spring 2005
Examination #4 ANSWER KEY
1.
(8 pts.) A sample of HgCl2 weighing 9.41 g is dissolved in 32.75 g of ethanol,
C2 H5OH (K b = 1.20 °C/m). The boiling point elevation of the solution is 1.27 °C.
Is HgCl2 an electrolyte in ethanol? Briefly explain with the appropriate
mathematical deduction.
∆Tb = i Kb m ⇒ i = ∆Tb /[Kb m] = 1.27 °C/[1.20 °C/m x 1.058 m] = 1.00
HgCl2 is NOT an electrolyte in solution as the van’t Hoff factor i = 1.
2.
(8 pts.) What is the boiling point of a solution of NaCl in water if the solution
freezes at –0.93 °C? Assume complete dissociation of the solute.
Kb (water) = 0.52 °C/m; Kf (water) = 1.86 °C/m.
∆Tf = i Kf m ⇒ m = ∆Tf/[i Kf ] = 0.93 °C/[2 x 1.86 °C/m] = 0.25 m
∆Tb = i Kb m = 2 x 0.52 °C/m x 0.25 m = 0.26 °C
Therefore, the boiling point of the solution = 100.26 °C
3.
(10 pts.) The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A
solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C.
The boiling point of pure ethanol is 78.35 °C, and the Kb for ethanol is 1.20 °C/m.
What is the molecular formula of fructose?
mol C = 40.0 g x 1 mol C/12 g C = 3.33 mol C
mol H = 6.7 g = 6.7 mol H
mol O = 53.3 g O x 1 mol O/16 g O = 3.33 mol O
EF = CH2 O
∆Tb = i Kb m ⇒ mol = [∆Tb /Kb ] x kg solvent = [0.24 °C/1.20 °C/m] x 0.325
kg
MM (g/mol) = 11.7 g/0.065 mol = 180 g/mol
Therefore, the molecular formula is C6 H12 O6
4.
(9 pts.) The osmotic pressure of a solution that contains 7.0 g of insulin per liter is
23 torr at 25 °C. What is the molecular mass of insulin?
π = MRT ⇒ 23 torr/760 torr atm-1 = 7.0 g/L x 1/MM x 0.08206 x 298
Therefore, the Molecular Mass = 5.7 x 103 g/mol
5.
(9 pts.) Lauryl alcohol is obtained from coconut oil and is used to make
detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes
at 4.1 °C. What is the molar mass of lauryl alcohol?
Kf (benzene) = 5.12 °C/m; normal freezing point (benzene) = 5.5 °C
∆Tf = Kf m ⇒ m = ∆Tf/K f = 1.4 °C/[5.12 °C/m] = 0.27 m
mol = 0.27 m x 0.100 kg = 0.027 mol solute
Therefore, the molar mass = 5.00 g/0.027 mol ≈ 190 g/mol
6.
(15 pts. total; 3 pts. each) An aqueous solution made up of 32.47 g of iron(III)
chloride in 100.0 mL of solution has a density of 1.249 g/mL at 25 °C. Calculate
its:
A.
B.
C.
D.
E.
mass percent
molarity
molality
osmotic pressure assuming complete dissociation of the solute
freezing point (K f for water = 1.86 °C/m)
A.
B.
C.
mass percent = 100% (32.47 g FeCl3 /124.9 g solution) = 26.00%
molarity = (32.47 g FeCl3 /162.2 g mol-1 )/0.1 L = 2.001 M
m = (32.47 g FeCl3 /162.2 g mol-1 )/[(124.9 g soln – 32.47 g solute) x 1
kg/1000 g)] = 2.166 m
π = iMRT = 4 x 2.001 M x 0.08206 x 298 K
= 195.7 atm
∆Tf = i Kf m = 4 x 1.86 °C/m x 2.166 m = 16.12 °C
Therefore, the freezing point of the solution is –16.12 °C
D.
E.
7.
(6 pts.) List the following aqueous solutions in order of increasing boiling point
and briefly explain your ordering:
0.120 M C6 H12O6 ; 0.050 M LiBr; 0.050 M Zn(NO3 )2
0.120 M glucose
0.050 M LiBr
0.050 M Zn(NO3 )2
i=1
i=2
i=3
Therefore, 0.120 M
Therefore, 0.10 M
Therefore, 0.15 M
Increasing boiling point: 0.050 M LiBr < 0.120 M glucose < 0.050 M
Zn(NO3 )2
Greater concentration will elevate the boiling point of each respective
solution; hence, the ordering above.
8.
Conceptual Short Answers (20 pts. total; 4 pts. each)
Criticize the following statements. If correct, state why. If incorrect, also state
why and correct the statement. ALL INCORRECT STATEMENTS ARE
CORRECTED WITH BOLDFACED, ITALICIZED, AND UNDERLINED
WORDS BELOW!
A.
For all DILUTE aqueous solutions, molarity and molality are equal.
A concentrated solution has more solute present, hence affecting the
calculated molarity and molality. A dilute aqueous solution is
assumed to consist predominantly of water, which has a density of 1
g/mL.
B.
Seawater and fresh water have DIFFERENT boiling points.
Since boiling point elevation is an example of a colligative property,
and seawater contains the dissolution of various salts, one would
expect that the boiling point of seawater is higher than the boiling
point of fresh water. Therefore, their respective boiling points are
different.
C.
The freezing point of a 0.10 m CaCl2 solution is 1.5 times that of a 0.10 m
KCl solutio n.
All other factors being equal, the van’t Hoff factor of 0.10 m CaCl2 is i
= 3, and the van’t Hoff factor of 0.10 m KCl is i = 2. Therefore, the
former solution will contain a freezing point that is 1.5 higher than the
latter, not twice.
D.
An ideal solution occurs when all intermolecular forces of attraction are
the same, resulting in no net enthalpy change.
This statement is entirely TRUE!
9.
(10 pts.) Henry’s law constants for aqueous solutions at 25 ºC are 8.20 x 10-7
M/mm Hg for N2 and 1.62 x 10-6 M/mm Hg for O2 . Determine the solubility of
nitrogen in water under an atmospheric pressure of 760 mm Hg, assuming that air
is 80% N2 and 20% O2 .
P(N2 ) = ?(N 2 ) PT
= (80 g N2 /28 g mol-1 N2 )/[(80 g/28 g mol-1 )+(20 g O2 /32 g mol-1 )] 760
= 623 mm Hg
solubility = kH Pgas = (8.20 x 10-7 M/mm Hg N2 )(623 mm Hg) = 5.11 x 10-4 M
10.
(5 pts.) An aqueous solution containing 12.0% MgCl2 by mass has a density of
1.105 g/mL. What is the mole fraction of water in this solution?
12 g MgCl2 / 100 g solution = 12 g MgCl2 / 88 g H2 O
Convert both to mol:
12 g MgCl2 = 0.1261 mol MgCl2 ; 88 g H2 O = 4.889 mol H2 O
Therefore, ?(H2 O) = 4.889 mol / [4.889 mol + 0.1261 mol] = 0.975