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DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY
CHEMISTRY OF SOLUTIONS
202-NYB-05 15/16
TEST 1
29 FEBRUARY 2012
INSTRUCTOR: I. DIONNE
Answers
Print your name: _____________________________________________
INSTRUCTIONS:
Answer all questions in the space provided.
Duration of this test is 75 minutes.
No books or extra paper are permitted.
Answer the questions in ink in order to preserve the right to grieve.
In order to obtain full credit for your answers, you must clearly show your work. Answers to
problems involving calculations must be expressed to the correct number of significant figures and
proper units.
5. Calculators may not be shared. Programmable calculators are not permitted.
6. Your attention is drawn to the College policy on cheating. This policy will be enforced.
7. A Periodic Table with constants is provided.
1.
2.
3.
4.
Problem 1:
/2
Problem 6:
/3
Sig. Fig.:
/1
Problem 2:
/3
Problem 7:
/3
Units:
/1
Problem 3:
/3
Problem 8:
/3
Problem 4:
/3
Problem 9:
/3
Problem 5:
/3
Total:
/28
(2 marks)
PROBLEM 1
Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate (Na2CO3) is
added to 30.0 mL of 1.0 M sodium bicarbonate (NaHCO3).
Na2CO3:
3.0 M x 0.0700 L = 0.21 mol of Na2CO3 or 0.42 mol Na+
NaHCO3:
1.0 M x 0.0300 L = 0.030 mol of NaHCO3 or 0.030 mol Na+
Total Na+ = 0.42 + 0.030 = 0.45 mol
Total volume = 0.0700 L + 0.0300 L = 0.1000 L
[Na+] = 0.45 mol / 0.100 L = 4.5 M
4.5 M
Answer: _______________
(3 marks)
PROBLEM 2
A bottle of wine contains 12.5% ethanol by volume. The density of ethanol (C 2H5OH) is 0.789
g/mL and that of water is 1.00 g/mL. Calculate the concentration of ethanol in wine in terms
of molality. [MM C2H5OH = 46.068 g/mol, MM H2O = 18.016 g/mol]
12.5% ethanol by volume implies
12.5 mL ethanol
12.5 mL ethanol + 87.5 mL water
12.5 mL ethanol x 0.789 g/mL = 9.8625 g ethanol
9.8625 g / 46.068 g/mol = 0.2140857 mol ethanol
87.5 mL water x 1.00 g/mL = 87.5 g water
87.5 g water = 0.0875 kg water
molality = 0.2140857 mol / 0.0875 kg = 2.4466937 m
2.45 m
Answer: _______________
Page |2
(3 marks)
PROBLEM 3
A 1.37 M solution of citric acid (H3C6H5O7) in water has a density of 1.10 g/cm3. Calculate the
mole fraction of the citric acid. [MM H3C6H5O7 = 192.124 g/mol, MM H2O = 18.016 g/mol]
1.37 M citric acid implies
1.37 mol citric acid
1 liter solution
1000 mL solution x 1.10 g/ml = 1100 g solution
1.37 mol citric acid x 192.124 g/mol = 263.20988 g citric acid
1100 g solution – 263.20988 g = 836.79012 g water
836.79012 g / 18.016 g/mol = 46.447054 mol water
mole fraction of citric acid = mol citric acid / total mol
mole fraction = 1.37 / (1.37 + 46.447054) = 0.0286509
0.029
Answer: _______________
PROBLEM 4
(3 marks)
How many grams of NaBr must be added to 250. g of water to lower the vapor pressure by
1.30 mmHg at 40°C assuming complete dissociation? The vapor pressure of water at 40°C is
55.3 mmHg. [MM NaBr= 102.89 g/mol, MM H2O = 18.016 g/mol]
Psolution = Χsolvent x Posolvent
Psolution = 55.3 mmHg – 1.30 mmHg = 54.0 mmHg
Χsolvent = 54.0 mmHg / 55.3mmHg = 0.97649
Χsolvent = 0.97649 = mol H2O / (mol H2O + mol Na+ and Br-)
mol H2O = 250 g / 18.016 g/mol = 13.877 mol H2O
Χsolvent = 0.97649 = 13.877 mol H2O / (13.877 mol H2O + mol Na+ and Br-)
(13.877 mol H2O + mol Na+ and Br-) = 14.211
mol Na+ and Br- = 0.334
But we have 2 moles of ions for every mol of NaBr; we have 0.167 mol NaBr
0.167 mol NaBr x 102.89 g/mol = 17.18 g
20 g
Answer: _______________
Page |3
(3 marks)
PROBLEM 5
(a) Which pairs of liquids will be soluble in each other?
yes
no
1. H2O and CH3CH2CH2CH3
2. C6H6 (benzene) and CCl4
3. H2O and CH3COOH
(b) A carrot is placed in a concentrated salt solution. What will most likely happen? (circle
your choice). Solvent flows from low solute concentration to high solute concentration
1. Water will flow from the carrot to the solution.
2. Water will flow from the solution to the carrot.
3. Salt will flow into the carrot.
4. Salt will precipitate out.
(c) Arrange the following aqueous solutions in order of increasing boiling points:
0.02 m LiBr, 0.03 m sucrose, 0.03 m MgSO4, 0.03 m CaCl2, and 0.025 m (NH4)2Cr2O7.
__________
< __________
< __________
< __________
sucrose
LiBr
CaCl2
MgSO4
(NH4)2Cr2O7 < __________
0.02 m LiBr = 0.04 m ions
more particles = higher boiling point
0.03 m sucrose = 0.03 m ions
0.03 m MgSO4 = 0.06 m ions
0.03 m CaCl2 = 0.09 m ions
0.025 m (NH4)2Cr2O7 = 0.075 m ions
Page |4
(3 marks)
PROBLEM 6
An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain
temperature by the reaction:
3H2(g) + N2(g) ⇌ 2NH3(g)
At equilibrium, the concentrations are [H2] = 5.0 M, [N2] = 8.0 M, and [NH3] = 4.0 M. What
were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?
3H2(g)
+ N2(g)
⇌ 2NH3(g)
I
initial
initial
0
C
−3x
−x
+2x
E
5.0
8.0
4.0
For NH3:0 + 2x = 4.0
x = 2.0 = [NH3]0
For H2: [H2]o −3x = 5.0
[H2]o = (5.0) + (3)(2.0) = 11.0
For N2: [N2]o −x = 8.0
[N2]o = 8.0 + 2.0 = 10.0
Answers: [N2]o = _______________
and [H2]0 _______________
11.0 M
10.0 M
Page |5
(3 marks)
PROBLEM 7
Exactly 4 mol sulfur trioxide is sealed in a 5.0-L container at 1529 K. Kp is 1150 for:
2SO3(g) ⇌ 2SO2(g) + O2(g)
Calculate the concentrations of all species at equilibrium.
[SO3]o = 4 mol / 5.0 L = 0.80 M
2SO3(g)
⇌ 2SO2(g) + O2(g)
I
0.80
0
0
C
−2x
+2x
+x
E
0.80 − 2x
0 + 2x
0 + x
K = Kp(RT)−Δn
K = 1150(0.08206 L·atm·K−1·mol−1 x 1529 K)−(3−2) = 9.16556
K = (2x)2(x) / (0.80 − 2x)2 ≈ (2x)2(x) / (0.80)2
x = 1.136 approximation is invalid
Assume that x = 0.35 M
[SO3]eq = 0.80 − (2)(0.35) = 0.10 M
[SO2]eq = (2)(0.35) = 0.70 M
[O2]eq = x = 0.35 M
0.35 M
0.70 M
0.10 M
Answers: [SO3]eq = ____________
, [SO2]eq = ____________
, [O2]eq = ____________
Page |6
(3 marks)
PROBLEM 8
Choose (circle) the one alternative that best answers the following questions.
(a) Consider the following equilibria involving SO2(g) and their corresponding equilibrium
constants.
SO2(g) + ½O2(g) ⇌ SO3(g)
2SO3(g) ⇌ 2SO2(g) + O2(g)
K1
K2
Which of the following expressions relates K1 to K2? (Circle your choice).
1. K2 = K12
2. K22 = K1
3. K2 = K1
4. K2 = 1/K1
5. K2 = 1/K12
(b) The reaction quotient for a reaction has a value of 75 while the equilibrium constant has a
value of 195. Which of the following statements is accurate?
1. The reaction must proceed to the left to establish equilibrium.
2. The reaction must proceed to the right to establish equilibrium.
3. The concentrations of the products will be much smaller that the
concentrations of the reactants when the system is at equilibrium.
4. The concentrations of the products will be about the same as the
concentrations of the reactants when the system is at equilibrium.
(c) Which of the following is not a colligative property?
1. osmotic pressure
2. solubility
3. boiling point elevation
4. freezing point depression
Page |7
(3 marks)
PROBLEM 9
(a) Consider the reaction:
Fe3+(aq) + SCN−1(aq) ⇌ FeSCN2+(aq)
How will the equilibrium position shift if
Shift left No change Shift right
←
→
1. water is added, doubling the volume?
2. AgNO3(aq) is added? (AgSCN is insoluble).
3. NaOH(aq) is added? [Fe(OH)3 is insoluble].
4 Fe(NO3)3(aq) is added?
accept
(b) This represents a system at equilibrium:
O2(aq) ⇌ O2(g)
In which direction will the equilibrium be shifted
Shift left No change hift right
←
→
by an increase in pressure?
by an increase in temperature?
Page |8