Worked Solutions
Chapter 8
Question 9
Sulfur dioxide reacts with hydrogen sulfide according to the equation
SO2(g) + 2H2S(g) → 3S(s) + 2H2O(g)
What volume of sulfur dioxide would react completely with 140 cm3 of hydrogen sulfide, all
volumes being measured under the same conditions of temperature and pressure?
Answer:
SO2(g)
+
→
2H2S(g)
1 molecule
2 molecules

1 volume
2 volumes

70 cm3
140 cm3
3S(s)
+
2H2O(g)
Question 10
280 cm3 of methane reacts completely with oxygen according to the equation
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Assuming that all volumes are measured under the same conditions of temperature and
pressure, (a) what volume of oxygen is consumed in the reaction and (b) what volume of
carbon dioxide is formed?
Answer:
CH4(g)
+
2O2(g)
→
CO2(g)
+
1 molecule
2 molecules
1 molecule
1 volume
2 volumes
1 volume
280 cm3
560 cm3
280 cm3
2H2O(g)
a) 560 cm3 of oxygen are consumed
b) 280 cm3 of carbon dioxide are formed.
Question 11
How many atoms are there in (a) 1 mole of iron (b) 0.25 moles of gold (c) 5 moles of helium
(d) 0.2 moles of lead (e) 0.75 moles of zinc (f) 2 moles of argon?
Answer:
(a) 1 mole
= 6 X 1023 atoms of iron
(b) 1 mole
= 6 X 1023 atoms
0.25 moles
= 6 X 1023 X 0.25 atoms
= 1.5 X 1023 atoms of gold
1
(c) 1 mole
5 moles
= 6 X 1023 atoms
= 6 X 1023 X 5 atoms
= 3 X 1024 atoms of helium
(d) 1 mole
0.2 moles
= 6 X 1023 atoms
= 6 X 1023 X 0.2 atoms
= 1.2 X 1023 atoms of lead
(e) 1 mole
0.75 moles
= 6 X 1023 atoms
= 6 X 1023 X 0.75 atoms
= 4.5 X 1023 atoms of zinc
(f) 1 mole
2 moles
= 6 X 1023 atoms
= 6 X 1023 X 2 atoms
= 1.2 X 1024 atoms of argon.
Question 12
How many molecules are there in (a) 2 moles of nitrogen (b) 0.3 moles of carbon dioxide (c)
0.05 moles of water (d) 0.75 moles of methane (e) 20 moles of ammonia (f) 0.1 moles of
hydrogen?
Answer:
(a) 1 mole
2 moles
= 6 X 1023 molecules
= 6 X 1023 X 2 molecules
= 1.2 X 1024 molecules of nitrogen
(b) 1 mole
0.3 moles
= 6 X 1023 molecules
= 6 X 1023 X 0.3 molecules
= 1.8 X 1023 molecules of carbon dioxide
(c) 1 mole
0.05 moles
= 6 X 1023 molecules
= 6 X 1023 X 0.05 molecules
= 3 X 1022 molecules of water
(d) 1 mole
0.75 moles
= 6 X 1023 molecules
= 6 X 1023 X 0.75 molecules
= 4.5 X 1023 molecules of methane
(e) 1 mole
20 moles
= 6 X 1023 molecules
= 6 X 1023 X 20 molecules
= 1.2 X 1025 molecules of ammonia
(f) 1 mole
0.1 mole
= 6 X 1023 molecules
= 6 X 1023 X 0.1 molecules
= 6 X 1022 molecules of hydrogen
2
Question 13
Calculate how many atoms there are in each of the quantities indicated in question 12.
Answer:
(a) 1 molecule of N2 contains 2 atoms of nitrogen
1.2 X 1024 molecules of N2 contain 1.2 X 1024 X 2
= 2.4 X 1024 atoms of nitrogen
(b) 1 molecule of CO2 contains 3 atoms, 1 of carbon and 2 of oxygen
1.8 X 1023 molecules of CO2 contain 1.8 X 1023 X 3
= 5.4 X 1023 atoms
(c) 1 molecule of H2O contains 3 atoms, 2 of hydrogen and 1 of oxygen
3 X 1022 molecules of H2O contain 3 X 1022 X 3
= 9 X 1022 atoms
(d) 1 molecule of CH4 contains 5 atoms, 4 of hydrogen and 1 of carbon
4.5 X 1023 molecules of CH4 contain 4.5 X 1023 X 5 atoms
= 2.25 X 1024 atoms
(e) 1 molecule of NH3 contains 4 atoms, 3 of hydrogen and 1 of nitrogen
1.2 X 1025 molecules of NH3 contain 1.2 X 1025 X 4 atoms
= 4.8 X 1025 atoms
(f) 1 molecule of H2 contains 2 atoms of hydrogen
6 X 1022 molecules of H2 contain 6 X 1022 X 2 atoms
= 1.2 X 1023 atoms of hydrogen
Question 14
A sample of nitrogen gas contains 9 X 1021 molecules. How many moles is this?
Answer:
6 X 1023 molecules = 1 mole
9 X 1021 molecules = 9 X 1021 / 6 X 1023 moles
= 0.015 moles
Question 15
How many moles are there in samples of carbon dioxide gas containing (a) 2.4 X 1025
molecules (b) 1.5 X 1021 molecules (c) 9 X 1023 molecules?
Answer:
(a) 6 X 1023 molecules = 1 mole
2.4 X 1025 molecules = 2.4 X 1025 / 6 X 1023 moles
= 40 moles
(b) 6 X 1023 molecules = 1 mole
3
1.5 X 1021 molecules = 1.5 X 1021 / 6 X 1023 moles
= 0.0025 moles
23
(c) 6 X 10 molecules = 1 mole
9 X 1023 molecules = 9 X 1023 / 6 X 1023 moles
= 1.5 moles
Question 16
What is the volume in litres at s.t.p. of (a) 0.5 moles of oxygen gas (b) 0.025 moles of
nitrogen gas (c) 3 moles of hydrogen sulfide gas?
Answer:
(a) 1 mole
0.5 moles
= 22.4 l at s.t.p.
= 0.5 X 22.4 l at s.t.p.
= 11.2 l of oxygen gas
(b) 1 mole
= 22.4 l at s.t.p.
0.025 moles = 0.025 X 22.4 l at s.t.p.
= 0.56 l of nitrogen gas
(c) 1 mole
3 moles
= 22.4 l at s.t.p.
= 3 X 22.4 l at s.t.p.
= 67.2 l of hydrogen sulfide
Question 17
What is the volume in cm3 at s.t.p. of (a) 0.05 moles of hydrogen gas (b) 0.125 moles of
methane gas (c) 7 moles of ammonia gas?
Answer:
(a) 1 mole
0.05 moles
= 22.4 l = 22400 cm3 at s.t.p.
= 0.05 X 22400 cm3 at s.t.p.
= 1120 cm3 of hydrogen gas
(b) 1 mole
0.125 moles
= 22.4 l = 22400 cm3 at s.t.p.
= 0.125 X 22400 cm3 at s.t.p.
= 2800 cm3 of methane gas
(c) 1 mole
7 moles
= 22.4 l = 22400 cm3 at s.t.p.
= 7 X 22400 cm3 at s.t.p.
= 156800 cm3 of ammonia gas
Question 18
How many moles are there in each of the following (all volumes measured at s.t.p.)?
(a) 44.8 l of carbon dioxide gas (b) 1.12 l of ammonia gas (c) 560 cm3 of oxygen gas
(d) 1120 cm3 of hydrogen gas (e) 56 cm3 of methane gas?
4
Answer:
(a) 22.4 l at s.t.p. = 1 mole
44.8 l at s.t.p. = 44.8 / 22.4 moles
= 2 moles of carbon dioxide
(b) 22.4 l at s.t.p. = 1 mole
1.12 l at s.t.p. = 1.12 / 22.4 moles
= 0.05 moles of ammonia
(c) 22,400 cm3 at s.t.p. = 1 mole
560 cm3 at s.t.p. = 560 / 22,400 moles
= 0.025 moles of oxygen
3
(d) 22,400 cm at s.t.p. = 1 mole
1120 cm3 at s.t.p. = 1120 / 22,400 moles
= 0.05 moles of hydrogen
(e) 22,400 cm3 at s.t.p. = 1 mole
56 cm3 at s.t.p. = 56 / 22,400 moles
= 0.0025 moles of methane
Question 19
Calculate the relative molecular mass of each of the following:
(a) H2S (b) CH4 (c) NH3 (d) C4H10 (e) C6H12O6 (f) C12H22O11 (g) C2H5OH.
Answers:
(a) Mr(H2S) = (2 X 1 + 32) = 2 + 32 = 34
(b) Mr(CH4) = (12 + 4 X 1) = 12 + 4 = 16
(c) Mr(NH3) = (14 + 3 X 1) = 14 + 3 = 17
(d) Mr(C4H10) = (4 X 12 + 10 X 1) = 48 + 10 = 58
(e) Mr(C6H12O6) = (6 X 12 + 12 X 1 + 6 X 16) = 72 + 12 + 96 = 180
(f) Mr(C12H22O11) = (12 X 12 + 22 X 1 + 11 X 16) = 144 + 22 + 176 = 342
(g) Mr(C2H5OH) = (2 X 12 + 5 X 1 + 16 + 1) = 24 + 5 + 16 + 1 = 46
Question 20
What is the mass of (a) 4 moles of iron (b) 0.3 moles of carbon dioxide (c) 0.125 moles of
nitrogen (d) 0.75 moles of benzene (C6H6) (e) 0.5 moles of sodium hydroxide?
Answer:
(a) Ar(Fe) = 56
Mass of 1 mole of iron = 56 g
Mass of 4 moles of iron = 4 X 56 g
= 224 g
5
(b) Mr(CO2) = 44
Mass of 1 mole of carbon dioxide = 44 g
Mass of 0.3 moles of carbon dioxide = 0.3 X 44 g
= 13.2 g
(c) Mr(N2) = 28
Mass of 1 mole of nitrogen = 28 g
Mass of 0.125 moles of nitrogen = 0.125 X 28 g
= 3.5 g
(d) Mr(C6H6) = 78
Mass of 1 mole of benzene = 78 g
Mass of 0.75 moles of benzene = 0.75 X 78 g
= 58.5 g
(e) Mr(NaOH) = 40
Mass of 1 mole of sodium hydroxide = 40 g
Mass of 0.5 moles of sodium hydroxide = 0.5 X 40 g
= 20 g
Question 21
How many moles in (a) 11 g of carbon dioxide (b) 196 g of sulfuric acid (c) 90 g of ethane
(C2H6) (d) 10 g of ammonium nitrate (NH4NO3) (e) 30 g of ethanoic acid (CH3COOH)?
Answer:
(a) Mr(CO2) = 44
44 g = mass of 1 mole of carbon dioxide
11 g = mass of 11 / 44 moles
= 0.25 moles.
(b) Mr(H2SO4) = 98
98 g = mass of 1 mole of sulfuric acid
196 g = mass of 196 / 98 moles
= 2 moles.
(c) Mr(C2H6) = 30
30 g = mass of 1 mole of ethane
90 g = mass of 90 / 30 moles
= 3 moles.
(d) Mr(NH4NO3) = 80
80 g = mass of 1 mole of ammonium nitrate
10 g = mass of 10 / 80 moles
= 0.13 moles.
6
(e) Mr(CH3COOH) = 60
60 g = mass of 1 mole of ethanoic acid
30 g = mass of 30 / 60 moles
= 0.5 moles.
Question 22
How many molecules are there at s.t.p. in (a) 280 cm3 of nitrogen gas (b) 2.24 l of oxygen
gas?
Answer:
(a) 22,400 cm3 at s.t.p. = 1 mole
280 cm3 at s.t.p. = 280 / 22,400 moles
= 0.0125 moles
1 mole = 6 X 1023 molecules
0.0125 moles = 0.0125 X 6 X 1023 molecules
= 7.5 X 1021 molecules.
(b) 22.4 l at s.t.p. = 1 mole
2.24 l at s.t.p. = 2.24 / 22.4 moles
= 0.1 moles
1 mole = 6 X 1023 molecules
0.1 moles = 0.1 X 6 X 1023 molecules
= 6 X 1022 molecules.
Question 23
How many atoms are there at s.t.p. in (a) 560 cm3 of methane (CH4) (b) 1.12 l of butane
(C4H10)?
Answer:
(a) 22,400 cm3 at s.t.p. = 1 mole
560 cm3 at s.t.p. = 560 / 22,400 moles
= 0.025 moles
1 mole = 6 X 1023 molecules
0.025 moles = 0.025 X 6 X 1023 molecules
= 0.15 X 1023 molecules
= 5 X 0.15 X 1023 atoms (since there are five atoms in
each methane molecule)
= 0.75 X 1023 atoms
= 7.5 X 1022 atoms.
7
(b) 22.4 l at s.t.p. = 1 mole
1.12 l at s.t.p. = 1.12 / 22.4 moles
= 0.05 moles
1 mole = 6 X 1023 molecules
0.05 moles = 0.05 X 6 X 1023 molecules
= 0.3 X 1023 molecules
= 14 X 0.3 X 1023 atoms (since there are fourteen atoms in
each butane molecule)
= 4.2 X 1023 atoms.
Question 24
What is the volume in litres at s.t.p. of (a) 2 g of methane gas (CH4) (b) 60 g of hydrogen
fluoride gas (HF) (c) 15 g of ethane gas (C2H6) (d) 10.5 g of propene gas (C3H6) (e) 80 g of
argon gas (Ar)?
Answer:
(a) Mr(CH4) = 16
16 g CH4 = 1 mole
2 g CH4 = 2 / 16 moles
= 0.125 moles
1 mole = 22.4 l at s.t.p.
0.125 moles = 0.125 X 22.4 l
= 2.8 l
(b) Mr(HF) = 20
20 g HF = 1 mole
60 g HF = 60 / 20 moles
= 3 moles
1 mole = 22.4 l at s.t.p.
3 moles = 3 X 22.4 l
= 67.2 l
(c) Mr(C2H6) = 30
30 g C2H6 = 1 mole
15 g C2H6 = 15 / 30 moles
= 0.5 moles
1 mole = 22.4 l at s.t.p.
0.5 moles = 0.5 X 22.4 l
= 11.2 l
8
(d) Mr(C3H6) = 42
42 g C3H6 = 1 mole
10.5 g C3H6 = 10.5 / 42 moles
= 0.25 moles
1 mole = 22.4 l at s.t.p.
0.25 moles = 0.25 X 22.4 l
= 5.6 l
(e) Ar(Ar) = 40
40 g Ar = 1 mole
80 g Ar = 80 / 40 moles
= 2 moles
1 mole = 22.4 l at s.t.p.
2 moles = 2 X 22.4 l
= 44.8 l
Question 25
How many molecules in (a) 180 g of water vapour (H2O) (b) 5 g of nitrogen monoxide (NO)
(c) 10 g of heptane (C7 H16) (d) 2.9 g of butane (C4 H10) (e) 1.5 g of methanal (HCHO)?
Answer:
(a) Mr(H2O) = 18
18 g = mass of 1 mole of water
180 g = mass of 180 / 18 moles
= 10 moles.
1 mole = 6 X 1023 molecules
10 moles = 10 X 6 X 1023 molecules
= 6 X 1024 molecules
(b) Mr(NO) = 30
30 g = mass of 1 mole of nitrogen monoxide
5 g = mass of 5 / 30 moles
= 0.166 moles.
1 mole = 6 X 1023 molecules
0.166 moles = 0.166 X 6 X 1023 molecules
= 1 X 1023 molecules
(c) Mr(C7 H16) = 100
100 g = mass of 1 mole of heptane
10 g = mass of 10 / 100 moles
= 0.1 moles.
9
1 mole = 6 X 1023 molecules
0.1 moles = 6 X 1022 molecules
(d) Mr(C4 H10) = 58
58 g = mass of 1 mole of butane
2.9 g = mass of 2.9 / 58 moles
= 0.05 moles.
1 mole = 6 X 1023 molecules
0.05 moles = 0.05 X 6 X 1023 molecules
= 3 X 1022 molecules
(e) Mr(HCHO) = 30
30 g = mass of 1 mole of methanal
1.5 g = mass of 1.5 / 30 moles
= 0.05 moles.
1 mole = 6 X 1023 molecules
0.05 moles = 0.05 X 6 X 1023 molecules
= 3 X 1022 molecules.
Question 26
What is the mass in g at s.t.p. of (a) 140 cm3 of hydrogen sulfide gas (H2S) (b) 560 cm3 of
hydrogen gas (H2) (c) 67.2 l of oxygen gas (O2) (d) 1.12 l of nitrogen dioxide gas (NO2) (e)
3.36 l of hydrogen chloride gas (HCl)?
Answer:
(a) 22,400 cm3 of hydrogen sulfide gas at s.t.p. = 1 mole
140 cm3 of hydrogen sulfide gas at s.t.p. = 140 / 22,400 moles
= 0.00625 moles
Mr(H2S) = 34
1 mole H2S = 34 g
0.00625 moles H2S = 0.00625 X 34 g
= 0.21 g
(b) 22,400 cm3 of hydrogen gas at s.t.p. = 1 mole
560 cm3 of hydrogen gas at s.t.p.
= 560 / 22,400 moles
= 0.025 moles
Mr(H2) = 2
1 mole H2 = 2 g
0.025 moles H2 = 0.025 X 2 g
= 0.05 g
(c) 22.4 l of oxygen gas at s.t.p. = 1 mole
10
67.2 l of oxygen gas at s.t.p.
= 67.2 / 22.4 moles
= 3 moles
Mr(O2) = 32
1 mole O2 = 32 g
3 moles O2 = 3 X 32 g
= 96 g
(d) 22.4 l of nitrogen dioxide gas at s.t.p. = 1 mole
1.12 l of oxygen gas at s.t.p.
= 1.12 / 22.4 moles
= 0.05 moles
Mr(NO2) = 46
1 mole NO2 = 46 g
0.05 moles NO2 = 0.05 X 46 g
= 2.3 g
(e) 22.4 l of hydrogen chloride gas at s.t.p. = 1 mole
3.36 l of oxygen gas at s.t.p.
= 3.36 / 22.4 moles
= 0.15 moles
Mr(HCl) = 36.5
1 mole HCl = 36.5 g
0.15 moles HCl = 0.15 X 36.5 g
= 5.48 g
Question 27
What is the volume at s.t.p. of a definite mass of gas that occupies a volume of 530 litres
at 20 0C and 102,000 Pa?
Answer:
P1V1
----- =
T1
P2V2
-----T2
P1 = 102000 Pa
P2 = 101325 Pa
V1 = 530 l
V2 = ? l
0
T1 = 20 C
T2 = 273 K
= 20 + 273 K
= 293 K
102000 X 530 = 101325 X V2
--------------------------------293
273
11
V2 = 102000 X 530 X 273
---------------------------293 X 101325
= 497.11 litres
Question 28
What is the volume at s.t.p. of a definite mass of oxygen gas that occupies a volume
of 560 cm3 at 30 0C and 100500 Pa?
Answer:
P1V1
----- =
T1
P1 = 100500 Pa
3
P2V2
-----T2
P2 = 101325 Pa
V1 = 560 cm
V2 = ? cm3
T1 = 30 0C
T2 = 273 K
= 30 + 273 K
= 303 K
100500 X 560 = 101325 X V2
--------------------------------303
273
V2 = 100500 X 560 X 273
---------------------------303 X 101325
= 500.45 cm3
Question 29
The volume of hydrogen collected in a reaction between zinc and hydrochloric acid was 240
cm3, measured at the laboratory conditions of 18 0C and 101000 Pa. Calculate the volume of
hydrogen at s.t.p.
Answer:
P1V1
----- =
T1
P2V2
-----T2
P1 = 101000 Pa
P2 = 101325 Pa
V1 = 240 cm3
V2 = ? cm3
T1 = 18 0C
T2 = 273 K
= 18 + 273 K
= 291 K
12
101000 X 240 = 101325 X V2
------------------
----------------
291
273
V2 = 101000 X 240 X 273
---------------------------291 X 101325
= 224.43 cm3
Question 30
What is the volume at 819 K and 100,000 Pa of a fixed mass of gas that occupies a volume of
5 litres at 91 K and 200,000 Pa?
Answer:
P1V1
----T1
P2V2
=
-----T2
P1 = 200 000 Pa
P2 = 100 000 Pa
V1 = 5 l
V2 = ? l
T1 = 91 K
T2 = 819
200 000 X 5
----------------
100 000 X V2
=
91
----------------819
V2 = 200 000 X 5 X 819
-------------------------91 X 100 000
= 90 l.
Question 36
A definite mass of gas has a volume of 2 m3 at 300 K and 100,000 Pa. How many moles of
gas is this?
Answer:
PV = nRT
P = 100,000 Pa
V = 2 m3
R = 8.31 J K-1 mol-1
13
T = 300 K
n = amount of gas in moles = PV / RT = 100,000 X 2
----------------------8.31 X 300
= 80.22 moles
Question 37
A chemical engineer wants to store a gas produced during a process at a chemical plant at a
pressure of 100 000 Pa and a temperature of 20 0C. The process produces 2,000 litres of gas
per hour, measured at 500000 Pa and 160 0C.
(a) What volume will this amount of gas occupy under the storage conditions?
(b) How many moles of gas are being produced per hour?
Answer:
(a) P1V1
-----
P2V2
=
T1
P1 = 500 000 Pa
T2
P2 = 100 000 Pa
V1 = 2000 l
V2 = ? l
T1 = 433 K
500 000 X 2000
------------------- =
433
------
T2 = 293
100 000 X V2
---------------293
V2 = 500 000 X 2000 X 293
---------------------------433 X 100 000
= 6766.74 litres
(b) PV = nRT
P = 500 000 Pa
V = 2000 l = 2000 X 10-3 m3
R = 8.31 J K-1 mol-1
T = 160 0C = 433 K
n = amount of gas in moles = PV / RT = 500 000 X 2000 X 10-3
-------------------------8.31 X 433
=277.91 moles
14
Question 38
A mass of 8.4 g of a gas occupies a volume of 5 x 10-3 m3 at 27 0C and
1 x 105 Pa. Calculate the relative molecular mass of the gas.
Answer:
PV = nRT
P = 1 x 105 Pa
V = 5 X 10-3 m3
R = 8.31 J K-1 mol-1
T = 27 0C = 300 K
n = amount of gas in moles = PV / RT = 1 x 105 X 5 X 10-3
----------------------8.31 X 300
= 0.20 moles
0.2 moles of the gas have a mass of 8.4 g.
1 mole has a mass of 8.4 / 0.2 g = 42 g
Relative molecular mass = 42
Question 39
A mass of 6.5 g of a gas occupies a volume of 1 x 10-2 m3 at 330 0C and
1 x 105 Pa. Calculate the relative molecular mass of the gas.
Answer:
PV = nRT
P = 1 x 105 Pa
V = 1 X 10-2 m3
R = 8.31 J K-1 mol-1
T = 330 0C = 603 K
n = amount of gas in moles = PV / RT = 1 x 105 X 1 X 10-2
----------------------8.31 X 603
=0.199
0.199 moles of the gas have a mass of 6.5 g.
1 mole has a mass of 6.5 / 0.199 g = 32.66 g
Relative molecular mass = 32.66
Question 41(b)
In a suitable apparatus at 423 K, 100 cm3 of sulfur dioxide gas (SO2) (measured at s.t.p.), was
reacted with 100 cm3 of hydrogen sulfide gas (H2S) (measured at s.t.p.). The total volume
15
(when corrected to s.t.p.) after the reaction was found to be 150 cm3. The products of the
reaction were water vapour and solid sulfur. When the water vapour formed was removed, a
volume (when corrected to s.t.p.) of 50 cm3 of gas remained. This gas was found to be sulfur
dioxide.
(i) What is the combining volumes ratio of sulfur dioxide and hydrogen sulfide? Show how
you arrive at your answer.
(ii) Write a correct chemical equation for the reaction.
Answer:
(i)
Initial volume of SO2 = 100 cm3
Final volume of SO2 = 50 cm3
Reacted volume SO2 = 50 cm3
Initial volume of H2S = 100 cm3
Final volume of H2S = 0 cm3
Reacted volume H2S = 100 cm3
Combining volumes ratio = SO2 : H2S = 50 : 100 = 1 : 2
(ii)
The balanced equation is: SO2 + 2H2S  3S + 2H2O
Question 44
(a) What is the mass of one mole of (i) oxygen (O2) (ii) sulfur dioxide (SO2)?
Answer:
(i) One mole of oxygen has a mass of 32 g
(ii) One mole of sulfur dioxide has a mass of 64 g
(b) What is the mass of 0.25 moles of sulfur dioxide?
Answer:
One mole of sulfur dioxide has a mass of 64 g
0.25 moles of sulfur dioxide have a mass of 64 X 0.25 = 16 g
(c) How many moles of sulfur dioxide are there in 8 g of sulfur dioxide?
Answer:
64 g of sulfur dioxide = 1 mole
8 g of sulfur dioxide = 8/64 moles = 0.13 moles
(e) How many SO2 molecules are there in 8 g of sulfur dioxide?
Answer:
8 g of sulfur dioxide = 0.125 moles
1 mole = 6 X 1023 molecules
0.125 moles = 6 X 1023 X 0.125 = 7.5 X 1022 molecules
16
(f) What is the volume at s.t.p. of 8 g of sulfur dioxide?
Answer:
8 g of sulfur dioxide = 0.125 moles
1 mole of gas at s.t.p. = 22.4 l
0.125 moles at s.t.p. = 22.4 X 0.125 = 2.8 l
Question 45
(a) How many moles of fluorine (F2) are there in 9.5g of fluorine gas?
(b) What volume will this quantity of fluorine occupy at s.t.p.?
(c) How many molecules are there in 9.5g of fluorine?
Answer:
(a) Mr(F2) = 38
38 g = 1 mole of fluorine
9.5 g = 9.5/38 = 0.25 moles fluorine
(b) 1 mole = 22.4 l at s.t.p.
0.25 moles = 22.4 X 0.25 l = 5.6 l at s.t.p.
(c) 1 mole = 6 X 10 23 molecules
0.25 moles = 0.25 X 6 X 10 23 = 1.5 X 10 23 molecules
Question 46
(a)What is the molar mass of ammonia (NH3)?
(b) What is the mass of 280 cm3 of ammonia gas at s.t.p.?
(c) How many ammonia molecules are present in this quantity?
Answer:
(a) Mr(NH3) = (14 + 3 X 1) = 14 + 3 = 17
The molar mass of ammonia is 17 g
(b) 22,400 cm3 of ammonia gas at s.t.p. = 1 mole
280 cm3 of ammonia gas at s.t.p.
= 280 / 22,400 moles
= 0.0125 moles
Mr(NH3) = 17
1 mole NH3 = 17 g
0.0125 moles NH3 = 0.0125 X 17 g
= 0.213 g
(c) 1 mole = 6 X 1023 molecules
0.0125 moles = 6 X 1023 X 0.0125 molecules
= 7.5 X 1021 molecules
17
Question 47
What mass of chromium will contain the same number of atoms as 3 g of carbon?
Answer:
Ar(C) = 12
12 g carbon = 1 mole
3 g carbon = 3 / 12 moles = 0.25 moles
0.25 moles of carbon contains the same number of atoms as 0.25 moles of chromium
1 mole of chromium = 52 g
0.25 moles chromium = 0.25 X 52 = 13 g
Question 48
(a) How many moles of carbonate ions are present in 5.3 g of anhydrous sodium carbonate
(Na2CO3)?
(b) How many moles of phosphate ions are present in 5 moles of calcium phosphate
(Ca3(PO4)2)?
Answer:
(a) Mr(Na2CO3) = 106
106 g = 1 mole
5.3 g = 5.3 / 106 moles = 0.05 moles Na2CO3
Since 1 mole of Na2CO3 contains 1 mole of carbonate ions, 0.05 moles Na2CO3 contains
0.05 moles of carbonate ions
(b) Since 1 mole of Ca3(PO4)2 contains 2 moles of phosphate ions, 5 moles Ca3(PO4)2
contains 10 moles of phosphate ions
Question 49
(a) What is the mass of 0.2 moles of nitrogen monoxide (NO) gas? How many molecules
are present?
(b) What volume at s.t.p. does this amount of nitrogen monoxide occupy?
(c) Given that nitrogen monoxide reacts with oxygen according to the equation
2NO(g) + O2(g) → 2NO2(g)
what volume of oxygen (measured at s.t.p.) is needed to react fully with this volume
of nitrogen monoxide?
Answer:
(a) Mr(NO) = 30
1 mole of NO = 30 g
0.2 moles of NO = 0.2 X 30 = 6 g
1 mole = 6 X 1023 molecules
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0.2 moles = 0.2 X 6 X 1023 molecules
= 1.2 X 1023 molecules
(b) 1 mole = 22.4 l at s.t.p.
0.2 moles = 0.2 X 22.4 l
= 4.48 l at s.t.p.
(c) According to the balanced equation, 2 moles of NO reacts with 1 mole of O2. Thus 0.2
moles of NO will react with 0.1 moles of O2.
1 mole = 22.4 l at s.t.p.
0.1 moles = 0.1 X 22.4 l at s.t.p.
= 2.24 l at s.t.p.
Question 50
(b) Convert each of the following to volumes at s.t.p.: (i) 37.5 cm3 of carbon dioxide at
300 K and 100,000 Pa (ii) 82 cm3 of chlorine at 18 0C and 102,000 Pa (iii) 225 l of
ammonia at 19 0C and 1,013,000 Pa (iv) 57 cm3 of chlorine at 100 0C and 99,000 Pa.
Answer:
(i)
P1 = 100 000 Pa
3
P2 = 101 325 Pa
V1 = 37.5 cm
V2 = ? cm3
T1 = 300 K
T2 = 273 K
100 000 X 37.5 = 101 325 X V2
---------------------------------300
273
V2 = 100 000 X 37.5 X 273
---------------------------300 X 101325
= 33.68 cm3 of carbon dioxide
(ii) P1 = 102 000 Pa
P2 = 101 325 Pa
V1 = 82 cm3
V2 = ? cm3
T1 = 18 0C
T2 = 273 K
= 18 + 273 K
= 291 K
102 000 X 82
-----------------291
= 101 325 X V2
---------------273
19
V2 = 102 000 X 82 X 273
--------------------------291 X 101 325
= 77.44 cm3 of chlorine
(iii) P1 = 1013 000 Pa
P2 = 101 325 Pa
V1 = 225 l
V2 = ? cm3
T1 = 19 0C
T2 = 273 K
= 19 + 273 K
= 292 K
1013 000 X 225 = 101 325 X V2
---------------------------------292
273
V2 = 1013 000 X 225 X 273
---------------------------292 X 101325
= 2103.08 litres of ammonia
(iv) P1 = 99 000 Pa
V1 = 57 cm3
T1 = 100 0C
P2 = 101 325 Pa
V2 = ? cm3
T2 = 273 K
= 100 + 273 K
= 373 K
99 000 X 57
= 101 325 X V2
---------------------------------373
273
V2 = 99 000 X 57 X 273
---------------------------373 X 101 325
= 40.76 cm3
Question 51
(c) In an experiment to find the relative molecular mass of a volatile liquid, a mass of
0.25 g of the volatile liquid was vapourised by heating it at 95 0C in a suitable
apparatus. If the volume of the vapour was 170 cm3 and the pressure was 1 x 105 Pa,
calculate the relative molecular mass of the volatile liquid.
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Answer:
PV = nRT
P = 1 x 105 Pa
V = 170 cm3 = 170 X 10-6 m3
R = 8.31 J K-1 mol-1
T = 95 0C = 368 K
n = amount of gas in moles = PV / RT = 1 x 105 X 170 X 10-6
----------------------8.31 X 368
= 0.00556
0.00556 moles of the gas have a mass of 0.25 g
1 mole has a mass of 0.25 / 0.00556 g = 44.96 g
Relative molecular mass = 44.96
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