Wolpa/AP Chemistry
Chapter 14: Solutions and Their Behavior
AP Problems
1. Concentrated sulfuric acid (18.4-molar H2SO4) has a density of 1.84 grams
per milliliter. After dilution with water to 5.20-molar, the solution has a density of
1.38 grams per milliliter and can be used as an electrolyte in lead storage
batteries for automobiles.
(a) Calculate the volume of concentrated acid required to prepare 1.00 liter of
5.20-molar H2SO4.
(b) Determine the mass percent of H2SO4 in the original concentrated solution.
(c) Calculate the volume of 5.20-molar H2SO4 that can be completely neutralized
with 10.5 grams of sodium bicarbonate NaHCO3.
(d) What is the molality of the 5.20-molar H2SO4 ?
ANSWER
(a) one point
V1M1 = V2M2
(1.00L) (5.20 mol/L) = (x) (18.4 mol/L)
x = 5.2 mol / (18.4 mol/L) = 0.283 L (or 283 mL)
(b) two points
mass 1 liter of concentrated H2SO4 = 1 L x (1.84 g/mL) x (1,000 ml/L) = 1,840 g H2SO4
18.4 mol H2SO4 x 98.1 g/mol = 1,805 g H2SO4
mass percent H2SO4 = (1,805 g / 1,840 g) x 100 = 98.1%
(c) three points
Stoichiometric ratio of NaHCO3 to H2SO4 = 2:1
10.5 g NaHCO3 x (1 mol NaHCO3 / 84.0 g NaHCO3) = 0.125 mol NaHCO3
Since 1 mol H2SO4 reacts with 2 mol NaHCO3, 0.125 mol NaHCO3 reacts with 0.0625
mol H2SO4
0.0625 mol H2SO4 = V x M = (V) (5.20 M)
V = 0.0625 mol / (5.20 mol/L) = 0.0120 L (or 12.0 mL)
(d) three points
molality = moles solute / 1,000 g solvent = moles solute / 1 kg solvent
mass of 1 L of 5.20 M H2SO4 = 1 L x (1,000 mL / 1 L) x (1.38 g 1 mL) = 1,380 g
mass of H2SO4 in 1 L = (5.20 mol/L) (98.1 g.mol) = 510 g
mass of H2O in 1 L = 1,380 - 510 = 870 g
molality = (5.20 mol H2SO4 / 870 g) x (1,000 g / 1 kg) = 5.98 m
Note: no credit earned for 5.20 mol / 1.38 kg = 5.77 m
2. An unknown compound contains only the three elements C,H, and O. A pure
sample of the compound is analyzed and found to be 65.60 percent C and 9.44
percent H by mass.
(a) Determine the empirical formula of the compound.
(b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is
observed to freeze at a temperature 15.2 Celsius degrees below the normal
freezing point of pure camphor. Determine the molar mass and apparent
molecular formula of the compound. (The molal freezing-point depression
constant, Kf, for camphor is 40.0 kg-K-mol¯1.)
(c) When 1.570 grams of the compound is vaporized at 300 °C and 1.00
atmosphere, the gas occupies a volume of 577 milliliters. What is the molar mass
of the compound based on this result?
(d) Briefly describe what occurs in solution that accounts for the difference
between the results obtained in parts (b) and (c).
ANSWER
(a) Assume a 100 gram sample ( not necessary for credit ):
65.60g C x (1 mol C / 12.01 g C) = 5.462 mol C
9.44g H x (1 mol H / 1.0079 g H) = 9.366 mol H
mass O = [100 - (65.60 + 9.44)] = 24.96 g O
24.96 g O x (1 mol O / 15.9994 g O) = 1.560 mol O
C5.462H9.366O1.560 ---> C3.5H6.0O1.0 ---> C7H12O2
One point earned for determining moles of C and moles of H
One point earned for determining moles of O
One point earned for correct empirical formula
(b) m = T / Kf = 15.2 °C /40.0 K kg mol¯1 = 0.380 mol / kg
0.01608 kg x (0.380 mo / 1 kg) = 0.00611 mol
molar mass = 1.570 g/ 0.00611 mol = 257 g / mol
One point earned for determination of molarity
One point earned for conversion of molarity to molar mass
OR,
moles solute = (T x kg solvent) / Kf = 0.00611 mol (one point)
molar mass = 1.570 g / 0.00611 mol = 257 g / mol (one point)
OR,
molar mass = (mass x Kf) / (T x kg solvent) = 257 g / mol (two points)
empirical mass of C7H12O2 = 7(12) + 12(1) + 2(16) = 128 g/mol
128 g/mol = 1/2 molar mass ---> molecular formula = 2x ( empirical formula) ----->
molecular formula = C14H24O4 (one point)
One point earned if molecular formula is wrong but is consistent with empirical formula
and molar mass
No penalty for simply ignoring the van't Hoff factor
Only one point earned for part (b) if response indicates that T= (15.2 + 273) = 288 K
and molar mass = 13.6 g / mol
(c) n = (pV) / (RT) = [(1 atm) (0.577 L)] / [(0.0821 L atm mol°1 K°1) (573 K)] = 0.0123
mol (one point)
molar mass = mass of sample / moles in sample = 1.570 g / 0.0123 mol = 128 g/mol (one
point)
Only one point can be earned for part (c) if wrong value for R is used and/or T is not
converted from C to K
(d) The compound must form a dimer in solution, because the molar mass in solution is
twice that it is in the gas phase,
OR,
the compound must dissociate in the gas phase ( A (g) --> 2B (g)) because the molar
mass in the gas phase is half that it is in solution.
One point earned for a reference to either or both the ideas of dimerization and
dissociation,br> -No point earned for a " non - ideal behavior " argument
3. Elemental analysis of an unknown pure substance indicates that the percent
composition by mass is as follows:
Carbon - 49.02%
Hydrogen - 2.743%
Chlorine - 48.23%
A solution that is prepared by dissolving 3.150 grams of the substance in 25.00
grams of benzene, C6H6, has a freezing point of 1.12°C. (The normal freezing
point of benzene is 5.50°C and the molal freezing-point depression constant, Kf,
for benzene is 5.12 C°/molal.)
(a) Determine the empirical formula of the unknown substance.
(b) Using the data gathered from the freezing point depression method, calculate
the molar mass of the unknown substance.
(c) Calculate the mole fraction of benzene in the solution described above.
(d) The vapor pressure of pure benzene at 35°C is 150. millimeters of Hg.
Calculate the vapor pressure of benzene over the solution described above at
35°C.
ANSWER
a) two points
carbon: 49.02 / 12.01 = 4.081 g
hydrogen: 2.743 / 1.008 = 2.722 g
chlorine: 48.23 / 35.453 = 1.360 g
mole ratios: C/Cl = 3; H/Cl = 2; Cl/Cl = 1
empirical formula = C3H2Cl
b) three points
[delta]Tf = Kf m
4.38 °C = (5.12) (x / 0.025)
x = 0.0214 mol
3.150 g / 0.0214 mol = 147 g / mol
Note: the scoring standards has this equation rather than the above three lines:
The standards then show:
MM = [(5.12) (3.150) (1000)] / [(4.38) (25)] = 147
c) two points
mole fraction = moles benzene / total moles
C6H6 = 78.108
25.00 g / 78.108 = 0.32 mol
0.32 / (0.32 + 0.0214) = 0.94
d) two points
Psoln = Ppure x mole fraction
Psoln = (150) (0.94) = 141 mm Hg
4. The molecular formula of a hydrocarbon is to be determined by analyzing its
combustion products and investigating its colligative properties.
(a) The hydrocarbon burns completely, producing 7.2 grams of water and 7.2
liters of CO2 at standard conditions.
(b) Calculate the mass in grams of O2 required for the complete combustion of
the sample of the hydrocarbon described in (a).
(c) The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution
prepared by mixing 100. grams of CHCl3 and 0.600 gram of the hydrocarbon is 64.0 °C. The molal freezing-point depression constant of CHCl3 is 4.68 °C / molal
and its normal freezing point is -63.5 °C. Calculate the molecular weight of the
hydrocarbon.
(d) What is the molecular formula of the hydrocarbon? 2)
ANSWER
a) three points
7.2 g H2O ÷ 18.0 g/mol = 0.40 mol H2O
0.40 mol H2O x (2 mol H / 1 mol H2O) = 0.80 mol H
7.2 L CO2 ÷ 22.4 L/mol = 0.32 mol CO2
0.32 mol CO2 x (1 mol C / 1 mol CO2) = 0.32 mol C
OR
n = PV ÷ RT = [(1 atm) (7.2 L)] ÷ [(0.0821 L atm mol¯1 K1) (273 K)] = 0.32 mol CO2
0.80 mol H ÷ 0.32 = 2.5
0.32 mol C ÷ 0.32 = 1
2.5 x 2 = 5 mol H
1 x 2 = 2 mol C
empirical formula = C2H5
b) two points
mol O2 for combustion = mol CO2 + 1/2 mol H2O = 0.32 + 0.20 = 0.52 mol O2
0.52 mol O2 x 32 g/mol = 17 g O2
alternate approach for mol O2 from balanced equation
C2H5 + 13/4 O2 ---> 2 CO2 + 5/2 H2O
other ratio examples:
1, 6.5 ---> 4, 5
0.25, 1.625 ---> 1, 1.25
mol O2 = 0.40 mol H2O x (13/4 mol O2 / 5/2 mol H2O) = 0.52 mol O2
Note: starting moles of C2H5 = 0.16 mol C2H5
c) three points
MM stands for molar mass.
T = (Kf (g/MM)) / kg of solvent
0.5 °:C = ((4.68 °:C kg mol¯:1) x (0.60 g / MM)) / 0.1 kg
MM = (4.68 x 0.60) / (0.5 x 0.1) = 56 or 6 x 101
an alternate solution for (c)
molality = 0.5 °:C / (4.68 0.5 °:C/m) = 0.107 m
mol solute =( 0.107 mol / kg solvent) x 0.100 kg solvent = 0.0107 mol
MM = 0.60 g / 0.0107 mol = 56 or 6 x 101
d) one point
(56 g/mol of cmpd) / (29 g/mol of empirical formula) = 1.9 empirical formula per mol
OR
6 x 101 / 29 = 2.1
empirical formula times 2 equals molecular formula = C4H10
5. The formula and the molecular weight of an unknown hydrocarbon compound
are to be determined by elemental analysis and the freezing-point depression
method.
(a) The hydrocarbon is found to contain 93.46 percent carbon and 6.54 percent
hydrogen. Calculate the empirical formula of the unknown hydrocarbon.
(b) A solution is prepared by dissolving 2.53 grams of p-dichlorobenzene
(molecular weight 147.0) in 25.86 grams of naphthalene (molecular weight
128.2). Calculate the molality of the p-dichlorobenzene solution.
(c) The freezing point of pure naphthalene is determined to be 80.2 °C. The
solution prepared in (b) is found to have an initial freezing point of 75.7 °C.
Calculate the molal freezing-point depression constant of naphthalene.
(d) A solution of 2.43 grams of the unknown hydrocarbon dissolved in 26.7
grams of naphthalene is found to freeze initially at 76.2 °C. Calculate the
apparent molecular weight of the unknown hydrocarbon on the basis of the
freezing-point depression experiment above.
(e)What is the molecular formula of the unknown hydrocarbon?
ANSWER
a) two points
Assume a 100-gram sample of hydrocarbon
93.46 grams C x (1 mole C / 12.01 grams C) = 7.782 moles C
6.54 grams H x (1 mole H / 1.008 grams H) = 6.49 moles H
7.782 moles C / 6.49 moles H = 1.20
C1.20H1.00 = C6H5
b) two points
Molality = moles solute per kilogram solvent
m = (2.53 grams solute / 25.86 grams solute) x (1 mole solute / 147.0 grams solute) x
(1000 grams solute / 1 kg solvent) = 0.665 m
c) one point
Freezing point lowering = 80.2° - 75.7° = 4.5°
Molal freezing point depression constant = (4.5° / 0.665 molal solution)
= 6.8° lowering for 1 molal solution
d) three points
Freezing point lowering = 80.2° - 76.2° = 4.0°
(6.8 x kg. solvent/mole solute) x (1 / 4.0°) x (2.43 grams solute / 26.7 grams solvent) x
(1000 grams solvent / kg solvent)
= 154 grams solute / mole solute)
e) one point
Empirical unit weight (C6H5) = 77
Number of empirical units per mole= 154 / 77 = 2
molecular formula = (C6H5) x 2 = C12H10
6. (a) A solution containing 3.23 grams of an unknown compound dissolved in
100.0 grams of water freezes at -0.97 °C. The solution does not conduct
electricity. Calculate the molecular weight of the compound. (The molal freezing
point depression constant for H2O is 1.86 °C kg mole¯1.)
(b) Elemental analysis of this unknown compound yields the following
percentages by weight: H = 9.74%; C = 38.70%; O = 51.56%. Determine the
molecular formula of the compound.
(c) Complete combustion of a 1.05-gram sample of the compound with the
stoichiometric amount of oxygen gas produces a mixture of H2O(g) and CO2(g).
What is the pressure of this gas when it is contained in a 3.00-liter flask at 127
°C?
ANSWER
a) 5 points
[delta] Tf= Kfm
0.97 = 1.86 m
m = 0.52 moles solute / kg solvent
3.23 g solute / 100 g solvent = 32.3 g solute / kg solvent
Mol. Wt. = (32.3 g solute/kg solvent) / 0.52 mole solute/kg solvent = 62
b) 5 points
Per 100 g of compound:
9.74 mole H / 1.0 = 9.74
38.7 mole C / 12.0 = 3.22
51.56 mole O / 16.0 = 3.22
Emperical formula is CH3O (fwt = 31)
Molecular formula is (CH3O)2 or C2H6O2
c) 5 points
Combustion produces 5 moles gaseous products per 1 mole C2H6O2
moles C2H6O2 = 1.05 g / 62.0 g/mol = 0.0169 mole
moles gases = 5 (0.0169) = 0.0846
PV = nRT
P = [(0.0846) (0.0821) (400)] / 3.00 = 0.962 or 0.93 atmosphere
For the problems in Parts A and B, a maximum of 1 point per question was deducted for
errors of the following kinds: arithmetic, significant figures, unit labels. Numerical
answers without supporting work were valued at one-third credit.