University of KwaZulu-Natal, Westville Campus, Durban
Class Test 2: Friday, 3 April 2009
GENERAL PRINCIPLES OF CHEMISTRY, CHEM110
Duration: 45 mins
Internal Examiners:
Total marks: 25
L. Pillay, and G. Maguire
IMPORTANT: Complete this part immediately.
Student No:
Name:
Tutorial Venue:
Tutor:
INSTRUCTIONS:
1. Student should note that there is negative marking in the multiple choice
questions. If you do not know the answer, do not guess.
2. Answer ALL questions.
3. Calculators may be used but all working must be shown.
4. The pages of this question paper must not be unpinned.
5. Your answers must be written on the question paper in the spaces provided. The lefthand pages may be used for extra space or for rough work.
6. Marks will be deducted for the incorrect use of significant figures and the omission of
units.
7. You must write legibly in black or blue ink. Pencils and Tipp-Ex are not allowed.
8. A periodic table is provided.
Constants:
Planck’s constant: h = 6.63 × 10-34 J s-1
Speed of Light: c = 3.00 × 108 m s-1
Rydberg Constant: RH = 1.0973731 x 107 m or 2.18 x 10-18 J
Question 1 MCQ: Circle the correct answer.
A) 1.0 mole of RbOH reacts with 0.9 moles of HNO3 in a neutralisation reaction to produce
RbNO3 and water. The following reaction occurred:
RbOH (aq) + HNO3 (aq) → RbNO3 (aq) + H2O (l)
How many moles of RbNO3 are produced?
a) 1.0
b) 2.7
c) 3.0
d) 0.9
[1]
B) 25.6 mL of water was added to a solution containing 0.765 moles of CaCl2. The number
of moles in the solution has now:
a) increased
b) decreased
c) remained the same
d) insufficient information to answer
[1]
C) 10 mL of a standard solution was pipetted into a volumetric flask. This solution was
made up to 100 mL with water. The concentration of the final solution is:
a) unchanged
b) less concentrated
c) more concentrated
d) insufficient information to answer the question
2
[1]
Question 1(continued)
D) Which of the following colours of light has the lowest energy photons?
a) green
b) violet
c) blue
d) red
e) yellow
[1]
E) Which quantum number defines the shape of an orbital?
a) principle, n
b) angular orbital momentum, l
c) magnetic, ml
d) spin, ms
[1]
F) Among the following sets of quantum numbers which set does not
correspond to an electron in a zinc atom (in it’s ground state)?
a) n = 4, l = 0, ml = 0, ms = +½
b) n = 3, l = 0, ml = 0, ms = +½
c) n = 4, l = 2, ml = 2, ms = -½
d) n = 2, l = 1, ml = -1, ms = -½
[1]
G) Which of the following electron configurations is correct for bromine?
a) [Ar] 4s24p5
b) [Ar] 4s24d104p5
c) [Ar] 4s23d104p5
d) [Kr] 4s24d104p5
[1]
3
Question 1(continued)
H) A compound consisting of an element having a low ionization potential and a second
element having a high electronegativity is likely to have
a) covalent bonds.
b) metallic bonds.
c) coordinate covalent bonds.
d) ionic bonds.
[1]
I) In which compound does ionic bonding predominate?
a) LiBr
b) CO
c) H2O
d) SiC
[1]
J) Which element attains the argon core when forming a compound by the transfer of
electrons?
a) Ca
b) Ba
c) Mg
d) Al
[1]
4
Question 2
2.744 g of iron(iii)oxide was weighed into a volumetric flask and titrated with 36.42 mL of
1.02 M phosphoric acid. The following reaction occurred:
2H3PO4 (aq) + Fe2O3 (s) → 2 FePO4 (aq) + 3H2O (aq)
a) Calculate the molar mass of Fe2O3
[1]
Fe = 55.85 g/mol O = 16.00 g/mol MW = 159.7 g/mol b) Calculate the number of moles of Fe2O3
[1]
n = m/M n = 2.744g / 159.7 g/mol n = 0.01718 mols
c) Identify the limiting reagent
[1]
n (H3PO4) = cv n = 1.02 mol/L x 0.03642 L n = 0.03715 mols ratio 2:1 → 0.01857 mols Fe2O3 is limiting ( ½) (½) d) Calculate the mass of iron phosphate formed (the molecular weight of iron
phosphate is 150.8 g/mol)
[1]
m = nM m = 0.01718 mols x 150.8 g/mol m = 2.591 g
5
Question 3
Adipic acid is composed of 62.68% C, 8.712% H and 27.90% O. It has a molecular mass of
114.1 amu.
a) Calculate the empirical formula of this compound
[3.5]
C H m 62.68 8.712 O 27.9 n 5.22 8.712 1.74 ÷ 1.74 3 5.0069 1 emp formula 3 5 1 b) Calculate the molecular formula
[1.5]
MW of C3H5O = 57.06 amu Factor = 114.1 amu / 57.06 amu = 2 Molecular formula: C6H10O2 6
½ 1 ½
1.5 ½
1 Question 4
A diode laser gives off monochromatic infrared light, with a wavelength of 785 nm. What is
the frequency (in s-1)?
[1]
νλ = c ν = c/λ = [3.00 × 108 ms-1]/[785 × 10-9 m] = 3.82 × 1014 s-1
(½ mark)
(½ mark)
Question 5
Calculate the wavelength (in nm) of the photon which must be absorbed in order to excite
an electron from the ground state (n=1) to the first excited state (n=2).
[2]
Two methods:
Method 1.
E = hν = RH[(1/ni)2 - (1/nf)2]
1/λ =
[RH / hc] x [(1/ni)2 - (1/nf)2]
(½ mark)
1/λ =
1.097 x 107 x [(1/ni)2 - (1/nf)2]
(½ mark)
1/λ =
1.097 x 107 x [(1/1)2 - (1/2)2]
(½ mark)
λ = 122 nm
(½ mark)
OR
Method 2.
E = hν = -RH[(1/ni)2 - (1/nf)2] =
-[2.18 × 10-18 J] x [(1/1)2 - (1/2)2]
= 1.64 × 10-18 J
(½ mark)
(½ mark)
E = hc/λ
7
λ = hc/E = [6.63 × 10-34 Js-1] x [3.00 × 108 m/s]/[1.64 × 10-18 J] =
(½ mark)
-7
λ = 1.22 × 10
m = 122 nm
(½ mark)
Question 6
Write the ground state electron configuration and identify the number of unpaired electrons
in each of the following atoms/ions.
a) Nitrogen
1s22s22p3 - 3 unpaired electrons (1.5 marks) b) Sulfur
1s22s22p63s23p4 - 2 unpaired electrons (1.5 marks)
[3]
[25]
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