CHEMISTRY DEPARTMENT
UNIVERSITY OF BOTSWANA
CHE101 General Chemistry I
Test 1
ANSWERS
Tuesday 11 September 2012
Time allowed: 2 hours
___________________________________________________________________________
Surname: _____________________________ First name: __________________________
I.D. No.:______________________________ Group: _______________________
___________________________________________________________________________
WRITE ALL ANSWERS ON THIS QUESTION PAPER!
This paper has two parts:
•
For section A (20 multiple choice questions), enter the letter for the correct answer
to each question in the grid on page 7.
•
For the 4 questions in section B, use the space provided to write your answers.
You may detach the Periodic Table, but you may NOT write anything on it during the
test.
IMPORTANT: IT IS THE STUDENT’S RESPONSIBILITY TO REPORT IF ANY
PAGE IS MISSING IN THIS PAPER. THE PAPER HAS 13 PAGES
INCLUDING A BLANK PAGE (PAGE 8) AND THE PERIODIC
TABLE.
___________________________________________________________________________
Useful data:
Avogadro’s constant: NA = 6.022 x 1023 mol−1
___________________________________________________________________________
Do NOT write in this table!
MC
1
2
3
1
4
Total
SECTION A: 20 Multiple Choice Questions (20 × 3 marks = 60 marks)
Enter the letter for the correct answer for each question in the grid on page 7.
___________________________________________________________________
1.
A sample of matter that is uniform throughout, and cannot be separated into other
substances by physical means, but can be decomposed into other substances by chemical
means, is called a(n)

(a)
compound.
(b)
element.
(c)
heterogeneous mixture .
(d)
homogeneous mixture.
(e)
mixture of elements.
2.

3.

4.

Which one of the following properties is an extensive property?
(a)
colour
(b)
density
(c)
melting point
(d)
mass
(e)
solubility in water
A measuring cylinder with 25.0 mL of water had a mass of 128.65 g. A metal object
was placed in the cylinder and completely submerged in the water. The water level rose
to 36.4 mL, while the mass of cylinder with contents was now 195.34 g. What is the
density of the metal object?
(a)
1.83 g/mL
(b)
5.15 g/mL
(c)
5.37 g/mL
(d)
5.85 g/mL
(e)
17.1 g/mL
Convert a density of 6.5 g/cm3 to kg/m3.
(a)
6.5 ×10─3kg/m3
(b)
6.5 × 10─1kg/m3
(c)
6.5 kg/m3
(d)
6.5 ×103kg/m3
(e)
6.5 ×105kg/m3
2
5.

6.
Accuracy refers to
(a)
how close a measured value is to the calculated value.
(b)
how close a measured value is to the true value.
(c)
how close a measured value is to other measured values.
(d)
the number of decimal places in a measured value.
(e)
the number of significant figures in a measured value.
What is the number of significant figures in each of the following measurements?
(i) 7.99 × 10-2 mL

(a)
3 in (i); 3 in (ii); 2 in (iii)
(b)
3 in (i); 3 in (ii); 4 in (iii)
(c)
5 in (i); 4 in (ii); 2 in (iii)
(d)
3 in (i); 4 in (ii); 4 in (iii)
(e)
5 in (i); 5 in (ii); 4 in (iii)
(ii) 0.0572 kg
(iii) 1.004 g
7.
Which symbol – name combination for elements is notcorrect?

(a)
Ar = Arsenic
(b)
Bi = Bismuth
(c)
Co = Cobalt
(d)
Ni = Nickel
(e)
Se = Selenium
8.

Lithium reacts with a certain nonmetallic element X to form an ionic compound with the
formula Li2X. Element X is a diatomic gas at room temperature. Element X must be
(a)
chlorine.
(b)
fluorine.
(c)
oxygen .
(d)
nitrogen.
(e)
sulfur.
3
9.
What is the nuclear symbol for the species that contains 23 protons, 27 neutrons and 21
electrons?
(a)
50
(b)
50
(c)
50
(d)
50

(e)
50
10.
Which one of the following classification of elements is not correct?

Co2+
Sc2+
Ru2+
Sn2+
V2+
(a)
At (Z = 85) is a halogen.
(b)
Ge (Z = 32) is a metalloid.
(c)
In (Z = 49) is a transition metal.
(d)
Kr (Z = 36) is a noble gas.
(e)
Sr (Z = 56) is an alkaline earth metal
11.
Which pair of elements would you expect to show the greatest similarity in their
physical and chemical properties?

(a)
Ca, Sr
(b)
Cs, Ba
(c)
Ga, Ge
(d)
H, F
(e)
H, Li
12.

Element X has two naturally occurring isotopes: X – 6 with mass of 6.015 amu and
abundance of 10.2%, and X – 7 with mass of 7.016 amu. The atomic weight of element
X is
(a)
6.12 amu
(b)
6.52 amu
(c)
6.91 amu
(d)
7 amu
(e)
13.031 amu
4
13.
What are the coefficients required to balance the following reaction equation:
__ C6H14(l) + __ O2(g)  __ CO2(g) + __ H2O(g)
(a)
1, 9, 6, 7 respectively
(b)
1, 10, 6, 7 respectively
(c)
2, 19, 6, 7 respectively
(d)
2, 19, 6, 14 respectively

(e)
2, 19, 12, 14 respectively
14.
What are the molar masses of the following compounds:
(i) KAl(SO4)2.12H2O
(ii) (NH4)2HPO4
(a)
(i) = 258.20 g/mol , (ii) = 114.02 g//mol
(b)
(i) = 258.20 g/mol, (ii) = 132.06 g/mol
(c)
(i) = 276.21 g/mol, (ii) = 131.05 g/mol
(d)
(i) = 474.38 g/mol, (ii) = 114.02 g/mol

(e)
(i) = 474.38 g/mol, (ii) = 132.06 g/mol
15.
Calculate the number of moles of Zn(NO3)2 in 143.5 g of this compound.

(a)
0.7577 mol
(b)
0.8877 mol
(c)
1.000 mol
(d)
1.127 mol
(e)
1.320 mol
16.
Potassium chlorate on heating decomposes as follows:
2KClO3(s)  2KCl(s) + 3O2(g)
Assuming 100% yield, calculate the mass of oxygen formed when 3.28 g of potassium
chlorate is heated.

(a)
0.571 g
(b)
0.856 g
(c)
1.28 g
(d)
1.93 g
(e)
2.57 g
5
17.

18.
An organic compound contains 80.0% C and 20.0% H by mass. What is the empirical
formula of the compound?
(a)
CH
(b)
CH3
(c)
CH4
(d)
C3 H
(e)
C4 H
A compound has the empirical formula NPF2 and a molar mass of 415 g/mol. What is
the molecular formula of the compound?
(a)
NPF2
(b)
N2P2F4
(c)
N3P3F6
(d)
N4P4F8

(e)
N5P5F10
19.
What is the mass% of carbon in dimethylsulfoxide (C2H6SO)?

20.
(a)
15.37 %
(b)
19.66 %
(c)
25.00 %
(d)
30.75 %
(e)
39.33 %
GeF3H is formed from GeH4and GeF4in the combination reaction:
GeH4+ 3GeF44GeF3H
If the reaction yield is 92.6%, how many moles of GeF4are needed to produce 8.00 mol
of GeF3H?

(a)
2.16 mol
(b)
2.78 mol
(c)
3.24 mol
(d)
5.56 mol
(e)
6.48 mol
6
Answer grid for Section A – Multiple Choice Questions
Enter the letter for the correct alternative for each question in the grid below.
Question
Answer
1
A
2
D
3
D
4
D
5
B
6
B
7
A
8
C
9
E
10
C
11
A
12
C
13
E
14
E
15
A
16
C
17
B
18
E
19
D
20
E
7
SECTION B (40%)
Answer each question in the space provided. Show your WORK and REASONING in
each question; answers without logical calculations will NOT be given credit. In your
calculations, give the final answer to the correct number of significant figures.
___________________________________________________________________
Question 1
(10 marks)
Calculate the number of H atoms in each of the following:
(i)
25.5 g C3H8
(ii)
3.00 x 1022 molecules of C5H11OH
(iii) 1.52 mol of C8H6O4
(i)
Molar mass C3H8 = 44.10 g mol-1
(could be implied, see next step:1 mark)
No. of moles C3H8 = 25.5 g / 44.10 g mol-1 = 0.5782 mol
(1 mark)
No. of moles H atoms = 0.5782 mol ×
(1 mark)
= 1.735 mol
No. of H atoms = 1.735 mol × 6.022 × 1023 mol-1 = 1.04 × 1024
(1 mark)
(answer to 3 sign. fig.: 1 mark)
(i) No. of moles of H atoms = 0,5782 mol x 8 mol H/1 mol C 3H8 = 4.626 mol and
therefore No. of H atoms = 4.626 mol x 6.022 x 1023 mol-1 = 2.79 x1024
(ii)
No. of H atoms = 3.00 × 1022 C5H11OH molecules ×
= 3.60 × 1023
(1 mark)
(answer to 3 sign. fig.: 1 mark)
(iii) No. of mol H atoms = 1.52 mol C8H6O4 ×
= 9.12 mol
No. of H atoms = 9.12 mol × 6.022 × 1023 mol-1 = 5.49 × 1024
(1 mark)
(1 mark)
(answer to 3 sign. fig.: 1 mark)
(The calculations in (i) and (iii) can also be done in one overall step – award marks
accordingly. I suspect however that most students will have used the individual steps as
given in the marking scheme.)
8
Question 2
(a)
(10 marks)
Give the names for the following compounds:
(i)
Cu3(PO4)2.3H2O
copper(II) phosphate trihydrate
(½ ½ ½
(ii)
Ca(HCO3)2
calcium hydrogencarbonate
(½
(iii)
marks)
KMnO4
½
marks)
potassium permanganate
( ½½
(iv)
P4O6
marks)
tetraphosphorus hexoxide
½½
(b)
Give the formulas for the following compounds:
(i)
Hypochlorous acid
HClO (aq)
½
(ii)
½
Ammonium iron(III) sulphate
NH4Fe(SO4)2
½½½
(iii)
Aluminium nitrate
Al(NO3)3
½
(iv)
Sodium sulphite
Na2SO3
½
(v)
Dinitrogen pentoxide
½
½
N2O5
½ ½
9
Question 3
(10 marks)
Indicate if the following statements are true or false. If false, correct the statement.
(a)
If the molecular weight of a compound is 116 amu., then the mass of 1 mol of this
compound is 116 g.
Correct
(b)
(1 mark)
Isotopes are atoms with the same number of nucleons but different number of protons.
False
(1 mark)
Isotopes are atoms with the same number of protons but different number of neutrons.
(2 marks)
(c)
A substance which consists of molecules must be a compound.
False
(1 mark)
A substance which consists of molecules could be a compound(e.g. water) or an element
(e.g. the element oxygen occurs naturally as molecules).
(2 marks)
(d)
The coefficients in the balanced reaction equation N2(g) + 3H2(g)  2NH3(g) show
that 1 g of N2 is required to react with 3 g of H2.
False
(1 mark)
The coefficients in the balanced reaction equation ……… show that 1 mol of N2 is
required to react with 3 mol of H2.
(2 marks)
10
Question 4
(10 marks)
Calcium carbide, CaC2, reacts with water to form calcium hydroxide Ca(OH)2, and the
flammable gas acetylene, C2H2, according to the reaction equation
CaC2(s) + 2H2O(l)  Ca(OH)2(aq) + C2H2(g)
50.0 g of water is added to 50.0 g of calcium carbide.
(a)
Which is the limiting reactant?
No. of moles of H2O available = 50.0 g / 18.02 g mol-1 = 2.775 mol
(1 mark)
No. of moles CaC2 available = 50.0 g / 64.10 g mol-1 = 0.7800 mol
(1 mark)
If all H2O reacts, no. of moles CaC2 required
= 2.775 mol H2O ×
= 1.388 mol CaC2
(1 mark)
Less CaC2 is available, therefore, CaC2 is the limiting reactant (and H2O is in excess).
(1 mark)
OR
If all CaC2 reacts, no. of moles H2O required
= 0.7800 mol CaC2 ×
= 1.560 mol H2O
(1 mark)
More H2O is available, therefore, H2O is in excess and CaC2 is limiting.
(b)
(1 mark)
Assuming 100% yield, what is the mass of acetylene formed?
No. of moles C2H2 formed = 0.7800 mol CaC2 ×
= 0.7800 mol C2H2(1 mark)
Mass C2H2 formed = 0.7800 mol × 26.04 g mol-1 = 20.3 g
(1 mark)
(Answer to 3 sign. fig. = ½ mark)
(c)
What is the mass of excess reactant left after the reaction?
No. of moles H2O reacted = 0.7800 mol CaC2 ×
= 1.560 mol H2O (1 mark)
Mass H2O reacted = 1.560 mol × 18.02 g mol-1 = 28.11 g
(1 mark)
Mass H2O left = (50.0 – 28.11) g = 21.9 g
(1 mark)
(Answer to 1 dec. place / 3 sign. fig. = ½ mark)
11
12