MCQs Test # 1 (Chapter
# 1 …. Introduction to Chemistry)
1. To determine the molecular formula of glucose, its empirical formula is multiplied by:
2. The
prefix 1018
3. What
is the ratio of the weight of water formed to the weight of oxygen used in the
formation of water?
4. One
mole of carbon-12 has a mass of:
5. What
is the mass of one mole of iodine molecule?
6. Which
of the following is classified as a monoatomic element?
7. The
present scale of atomic mass is based on 1 amu being equal to the mass of:
8. The
mass of a single hydrogen atom is:
9. 11.2
dm3 of nitrogen gas at STP weighs:
10.The mass of one atom of coal is:
Explanatory Answers of MCQs Test # 1 (Chapter # 1 …… Introduction to Chemistry)
1. Solution
For glucose, value of integer n is 6.
2. Solution
Prefixes (Multiples & Fractions) used with S.I. and Metric Units
|
Prefixes |
Multiples |
Prefixes |
Fractions |
|
Deca (D) |
101 (Ten) |
deci (d) |
10–1 |
|
Hecto (H) |
102 (Hundred) |
centi (c) |
10–2 |
|
kilo (k) |
103 (Thousand) |
milli (m) |
10–3 |
|
Mega (M) |
106 (Million) |
micro (µ;
muo) |
10–6 |
|
Giga (G) |
109 (Billion) |
nano (n) |
10–9 |
|
Tera (T) |
1012 (Trillion) |
pico (p) |
10–12 |
|
Peta (P) |
1015 (quadrillion) |
femto (f/fm) |
10–15 |
|
Exa (E) |
1018 (quintillion) |
atto (a) |
10–18 |
|
Zetta (Z) |
1021 (sextillion) |
zepto (z) |
10–21 |
|
Yottta (Y) |
1024 (septillion) |
yacto (y) |
10–24 |
3. Solution
4. Solution
Mass of one mole of carbon-12 is 12 g or 12/1000 = 0.012 kg
5. Solution
6. Solution
Noble gases (e.g. He, Ne, Ar, Kr, Xe and Rn) are
classified as monoatomic elements.
7. Solution
An atomic mass unit (abbreviated as a.m.u) is a physiochemical constant and it is defined as one twelfth (1/12) of the mass of a single atom (the most abundant lightest isotope) of carbon-12 (12C).
[Because the mass of an atom’s electrons is negligible compared with the mass of its proton and neutrons, defining 1 amu as 1/12 the mass of a 12C atom means that both protons and neutrons have a mass of almost exactly 1 amu].
1 amu is equivalent to 1.66 x 10–24 g OR 1.66 x 10–27 kg (which is approximately equal to the mass of one hydrogen atom). Prior to 1960, the amu was defined in terms of the mass of a 16O-isotope (1.6599 x 10–27 kg).
8. Solution
9. Solution
22.4 dm3 of nitrogen gas at STP weighs 28 g
10. Solution
11. Solution
The weight of one mole of KAl(SO4)2.12H2O is equal to its gram formula mass as it is an ionic compound.
KAl(SO4)2.12H2O
= 39 + 27 + 2(32) + 8(16) + 12(18) = 474 g
12. Solution
13. Solution
14. Solution
32 g of oxygen is present in 1 mole of CO2
1 g of oxygen is present in 1/32 mole of CO2
8 g oxygen is present in 1/32 x 8 mole of CO2
= 0.25 mole of CO2
15. Solution
Mass of one electron is 9.11 x 10-31 kg
Mass of one mole (6.02 x 1023) electron
is = 9.11 x 0-31 x 6.02 x 1023 = 54.84 x 10‒8 kg or 5.5 x 10‒7
kg
Mass of one mole of electron in mg = 5.5 x 10-7
kg x 1 x 106 = 5.5
x 10‒1 mg or 0.55 mg
16. Solution
8 of O constitutes its half mole.
17. Solution
18. Solution
Protons in 1 mole of CaCO3 = Z of Ca + Z
of C + Z of O x 3 Þ 20 + 6 + 24 = 50 moles of
protons
100 g of CaCO3 contains 50 moles of
protons
1 g of CaCO3 contains 50/100 moles of
protons
10 of CaCO3 contains = 50/100 x 10 x
6.02 x 1023 moles of protons = 3.0115 x 1024 protons
19. Solution
2 g of oxygen constitutes its 0.125 mole
0.125 mole of oxygen contains number of atoms equal
to 0.125 x 6.02 x 1023 = 7.525 x 1022 atoms
0.5 g of hydrogen constitutes its 0.25 mole
2.3 g of sodium constitutes its 0.1 mole
7 g of nitrogen constitutes its 0.5 mole
4 g of sulphur constitutes its 0.125 mole
4 g of sulphur and 2 g of oxygen constitutes same
number of moles (i.e. 0.125 mole), therefore, they contain same number of
atoms.
20. Solution
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