School of Chemistry and
Physics
Westville Campus,
Durban
BARCODE
Test 2: Wednesday 9 April 2014
CHEM110: GENERAL PRINCIPLES OF CHEMISTRY
Duration: 45 minutes
Time: 17:45 – 18:30
Examiner: M. Shozi
IMPORTANT: Complete this part immediately.
Name
Total Marks: 25
MODEL ANSWER
Student No.
Tutorial Group
Tutorial Venue
Tutor’s Name
INSTRUCTIONS:
1.
Answer ALL questions.
2.
Calculators may be used but all workings must be shown.
3.
The pages of this paper must not be unpinned.
4.
Your answers must be written on the question paper in the spaces provided for section B
ONLY. The left hand facing pages may be used for extra space or for rough work.
5.
Marks will be deducted for the incorrect use of significant figures and the omission of units.
6.
You must write legibly in black or blue ink. Pencils and Tipp-ExTM are not allowed.
7.
This question paper consists of 8 pages excluding a periodic table and data sheet. Please
check that you have them all.
8.
A periodic table and a data sheet are provided.
Section A
Total
Maximum Mark
Mark
14
Section B
1
2
3
Total
4
4
3
11
Test
Total
25
School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 110: General Principles of Chemistry
TEST 2: Wednesday 9 April 2014
SECTION A - Multiple Choice Questions

For each of the following questions, select the correct answer from the list
provided.

There is only one correct answer for each question.

Indicate your answer on the multiple choice answer sheet provided.

Make a dark heavy mark with HB pencil that fills the block of the appropriate letter
completely.
1.
What is the oxidation number of sulfur in HSO4−?
A +2
B +4
C +6
D +8
2.
(1)
Identify the reducing agent in the following reaction:
Cu + 4 HNO3  Cu(NO3)2 + 2 NO2 + 2 H2O
A Cu
B HNO3
C Cu(NO3)2
D NO2
3.
(1)
Which of the following is a weak base?
A NaOH
B NH3
C CsOH
D Ca(OH)2
(1)
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School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 110: General Principles of Chemistry
TEST 2: Wednesday 9 April 2014
4.
A disproportionation reaction is one in which
A the equation is not balanced.
B one reactant is in greater proportion than the other.
C the same reactant is oxidised and reduced.
D water is added.
5.
(1)
What volume of 2.000 M HCl is needed to prepare 250.0 mL of 0.3000 M HCl?
A 1660 mL
B 37.50 mL
C 0.3000 mL
D 1.660 mL
6.
(1)
What is the concentration of OH− ions in a 0.5000 L solution of 0.2600 M Ca(OH)2?
A 0.1300 M
B 0.05200 M
C 0.01300 M
D 0.5200 M
(2)
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School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 110: General Principles of Chemistry
TEST 2: Wednesday 9 April 2014
7.
What volume of 0.1000 M SO32– is needed to titrate 24.00 mL of 0.2000 M Fe3+?
2 Fe3+(aq) + SO32–(aq) + H2O(l)  2 Fe2+(aq) + SO42–(aq) + 2 H+(aq)
A 48.00 mL
B 24.00 mL
C 12.00 mL
D 6.00 mL
8.
(2)
Boyle’s Law states that for a fixed amount of gas at constant temperature,
A its pressure increases as its volume increases.
B its pressure increases as its volume decreases.
C its pressure is equal to its volume.
D its pressure increases and its volume remains the same.
(1)
9.
One mole of O2(g)
A has a mass of 16.00 g.
B occupies about the same volume as a mole of Cl2(g) at STP.
C occupies 22.414 L at any conditions of temperature and pressure.
D contains two atoms.
(1)
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School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 110: General Principles of Chemistry
TEST 2: Wednesday 9 April 2014
10.
Ideal behaviour for a gas is most likely to be observed under conditions of
A low temperature and high pressure.
B low temperature and low pressure.
C high temperature and low pressure.
D high temperature and high pressure.
11.
(1)
What is the density (in g L-1) of N2 gas at 350 K and 0.921 atm?
A 1.11
B 8.87 x 10-3
C 0.899
D 113
(2)
[14]
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School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 110: General Principles of Chemistry
TEST 2: Wednesday 9 April 2014
SECTION B: Full Questions.
Answer on the question paper in the space
provided.
1.
Balance the following redox reaction in basic medium using the half-reaction
method. Show all relevant steps.
Cr(OH)3 + ClO−  CrO42− + Cl−
(4)
Oxidation half-reaction:
H2O + Cr(OH)3  CrO42− + 5 H+ + 3e- (x2)
(½ each for balancing O, H, charge)
Reduction half-reaction:
_
2e- + 2 H+ + ClO  Cl− + H2O (x3)
(½ each for balancing O, H, charge)
Overall reaction:
_
2 Cr(OH)3 + 3 ClO  2 CrO42− + 3 Cl− + 4 H+ + H2O
_
4 OH− + 2 Cr(OH)3 + 3 ClO  2 CrO42− + 3 Cl− + 4 H+ + H2O + 4 OH−
_
4 OH− + 2 Cr(OH)3 + 3 ClO  2 CrO42− + 3 Cl− + 5 H2O
(½ each for adding 4 OH− and for overall balanced reaction)
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School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 110: General Principles of Chemistry
TEST 2: Wednesday 9 April 2014
2.
To determine the purity of acetylsalicylic acid (C9H8O4, 180.2 g mol-1) in an aspirin
tablet, a 0.3000 g sample of aspirin was reacted with 42.78 mL of 0.1000 M NaOH
and the solution was heated to complete the reaction as shown below:
C9H8O4(s) + 2 NaOH(aq)  CH3COONa(aq) + HOC6H4COONa(aq) + H2O(l)
After the reaction mixture was cooled, the excess NaOH was back-titrated with
14.29 mL of 0.1056 M HCl according to the following reaction:
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
What is the percentage purity (% m/m) of acetylsalicylic acid in the aspirin sample?
(4)
Moles of NaOH added = 0.04278 L x 0.1000 M = 0.004278 mol ½
Moles of HCl used = 0.01429 L x 0.1056 M = 0.001509 mol ½
Therefore moles excess NaOH = moles HCl used = 0.001509 mol ½
Moles NaOH reacted with acetylsalicylic acid = 0.004278 – 0.001509 mol
= 0.002769 mol 1
Moles of acetylsalicylic acid = ½ x moles NaOH = 0.001385 mol ½
Mass acetylsalicylic acid = 0.001385 mol x 180.2 g mol-1 = 0.2496 g ½
% purity = (0.2496 g ÷ 0.3000 g) x 100 = 83.20 % ½
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School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 110: General Principles of Chemistry
TEST 2: Wednesday 9 April 2014
3.
Ammonium nitrite (NH4NO2, 64.05 g mol-1) decomposes on heating to form
nitrogen gas and water:
NH4NO2(s)  N2(g) + 2 H2O(l)
Decomposition of a sample of NH4NO2 produces 515.6 mL of N2 gas collected over
water at 299.2 K and 0.9800 atm. Calculate the mass of NH4NO2 which
decomposed. Vapour pressure of H2O at 299.2 K is 0.0329 atm.
(3)
From Dalton’s law of partial pressures:
.:
PN2 = PTotal - PH2O = 0.9800 – 0.0329 = 0.9471 atm ½
From Ideal gas law: 1½ (½ for equation and 1 for calculation)
i)
Using R = 0.08206 L atm K-1 mol-1
nN2 =
ii)
PV
RT
=
0.9471 atm x 0.5156 L
0.08206 L atm K
-1
-1
mol
x 299.2 K
= 1.989 x 10-2 mol
Using R = 8.314 kPa dm3 K-1 mol-1
nN2 =
PV
RT
=
95.96 kPa x 0.5156 dm3
8.314 kPa dm3 K-1 mol-1 x 299.2 K
= 1.989 x 10-2 mol
Moles NH4NO2 = moles N2 = 1.989 x 10-2 mol ½
Mass NH4NO2=n x MM= (1.989 x 10-2 mol)(64.05 g mol-1) = 1.274 g ½
[11]
END OF PAPER
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