Mole Concept Moles Equations and Molarity
Mole Concept Moles Equations and Molarity
Mole Concept Moles Equations and Molarity
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The mole
A mole of a substance is the mass in grams of the substance which is numerically equal to its
relative atomic mass or its relative molecular mass.
E.g. One mole of carbon weights 12g, 1 mole of oxygen molecule weights 32g. 1 mole of the
compound ammonium sulphate weights 132g.
- A mole of any substance contains the same number of particles. These particles can
be molecules, a toms, ions or electrons.
A mole of any substance contains 6.02 x 1023 particles. These number of particles in any
mole of a substance (6.02 x 1023) is called AVOGADRO’S NUMBER
There are 6.02 x 1023 carbon atoms in 1 mol (12g) of carbon. There are 6.02 x 1023 oxygen
molecules in 1 mol (32g) of oxygen. There are 6.02 x 1023 formula units of ammonium
sulphate in 1 mol (132g) of ammonium sulphate.
Example 1
Taking Avogadro’s constant equal to 6 x 1023
How many Cu atoms are there in a copper plate, weighing 48g [Cu = 64]
Solution:
Example 2
How many grams of Ag [Ag = 108] contain 1. 2 x 1023 atoms
Solution:
Let the number of moles in 1.2x1023 atoms be X
1 mole of Ag contain 6 x1023 atoms
X moles of Ag contain 1.2 x 1023 atoms
∴ 1.2 x1023 x 1
6x1023
= 0.2 moles
But 1 mole of Ag contain 108g
∴ 0.2 moles of Ag contain [108 x 0.2] g
1
= 21.6g
Example 3
How many C atoms are there in a carbon rod weighing 8 g (C = 12)
Solution
12g of carbon contains 6 x1023 atoms
8 𝑥 6 𝑥 1023
∴ 8g of carbon contain = 4 𝑥 1023 𝑎𝑡𝑜𝑚𝑠
12
Example 4
How many grams of copper (Cu = 64) contain 4.5 x 1023 atoms?
Solution
6 x1023 atoms of copper weigh 64g
4.5 𝑥 1023 𝑥 64
∴ 4.5 x 1023 atoms contain 6 𝑥 1023
= 48𝑔
Equations
If we consider the following equation:
C + O2 → CO2
The equation now may mean
Similarly
2C + O2 → 2CO
Example 5
What is the mass of
(a) 0.1 mole of CaSO4
R.F.M CaSO4 = 40 + 32 + 4 X 16 = 136g
=> 1 mole weighs 136g
0.1mole weigh 136 x 0.1 = 13.6g
Example 7:
According to the equation
2C + O2 → 2CO
How many moles of carbon will react with 0.2mol of oxygen?
Solution
From the equation
2 mol of C reacted with 1 mol of O2
=> 1 mole of O2 requires 2 moles of C
∴ 0.2 moles of O2 requires 2 x 02 = 0.4 moles of C
Example 8:
According to the equation
C3H8 + 502 → 3CO2 + 4H2O
(a) How many moles of CO2 will be produced in the reaction of 3.2g of O2?
Solution
Mass of 5 mole of oxygen molecules = 5 (2 x 16) = 160g
∴ 160g of oxygen produce 3 mole of carbon dioxide
3 𝑥 3.2
3.2g of oxygen produce 160 = 0.06𝑚𝑜𝑙𝑒𝑠
(b) How many grams of propane will react with 0.5moles of O2?
Solution
Formula mass of propane, C3H8 = 12 x 3 + 1 x 8 = 44g
5 moles of oxygen react with 1 mole of propane
1 𝑥 0.5
∴ 0.5 moles of oxygen react with 5 = 0.1 𝑚𝑜𝑙𝑒 of propane
2. Morality of a solution is the number of moles of the solute contained in 1 litre, 1dm3
or 1000cm3 of the solution.
3. A two molar solution of sodium hydroxide (2M NaOH) is a solution containing two
moles of the NaOH in 1000 cm3 of the solution
4. Some formulas
Morality = Concentration in gm/L (units for morality mol dm-3)
R.F.M
Mole = grams/mass
R.F.M
Example 1
20cm3 of 0.1M NaOH completely reacted with 50cm3 of dilute HCl. Calculate the Morality
of the acid and concentration of the acid in g/l
Solution
Reacting equation
Given: 20 cc 50cc
0.1M ?
0. 1
20cm3 of NaOH will contain x 20 moles
1000
Morality of HCl
50 cm3 of HCl contained 0.002 moles
0.002
1 cm3 of HCl will contain moles
50
0.002
1000 cm3 of HCl will contain x1000 moles = 0.04 M
50
B. 20.0 𝑥 0.25
2 𝑥 0.1
C. 2 𝑥 20.0 𝑥 0.25
0.1
D. 2 𝑥 20.0 𝑥 0 .1
0.25
B. 0.2 𝑥 20
2𝑥1
C. 0.2 𝑥 10
2 𝑥 20
D. 2 𝑥 0.2 𝑥 20
10
7 CaCO3 (s) heat CaO(s) + CO2(g)
The mass, in grams of calcium oxide formed when 20g of calcium carbonate
completely decomposes is
(Ca = 40, C = 12, O= 16)
A. 20 𝑥 56
100
B. 20 𝑥 100
56
C. 44 𝑥 56
100
D 20 𝑥 44
100
8. What mass, in grams, of sodium carbonate-10-water, Na2CO3.10H2O, is
contained in 50cm3 of a 0.1M solution?
A. 106 𝑥 0.1 𝑥 1000
50
B. 106 𝑥 0.1 𝑥 50
1000
D. 106 𝑥 0.1 𝑥 50
1000
12 The volume of 0.2M sodium hydroxide solution which neutralise 25cm3 0.1M
hydrochloric acid is
A. 5cm3 B. 12.5cm3 C. 25cm3 D. 50cm3
13 Which one of the following contains the same number of atoms as 8g of
sulphur?
A. 20g of calcium
B. 10g of calcium
C. 12g of carbon
D. 4 g of carbon
14 What mass of sulphuric acid (Mr 98) in 5cm3 of 0.2M sulphuric acid solution
A. 98 𝑥 5
0.2𝑥 1000
B. 98 𝑥 0.2 𝑥5
1000
C. 98 𝑥 0.2
5 𝑥 1000
D. 9.8 𝑥 5 1000
0.2
B. 0.00125 𝑥 1000
2 𝑥 25
C 0.00125 𝑥 2 𝑥 1000
22.7
D. 0.00125 𝑥 1000
25
B 44 𝑥 100
20
C. 20 𝑥 100
44
D. 20 𝑥 44
100
B. 0.64 𝑥 64
80
C. 0.64 𝑥 96
80
D. 0.64 𝑥 80
64
19 Lead (II) nitrate reacts with potassium iodide according to the following
equation
Pb(NO3)2 (aq) + 2KI (aq) PbI2(s) + 2KNO3(aq)
The mass pf lead (II) iodide formed when 33.2g of potassium iodide is reacted
with excess lead (II) nitrate is (K= 39, I= 127, Pb = 207)
A. 16 g B. 46.1g C. 66.4g D. 92.2g
22 Copper (II) sulphate reacts with sodium carbonate according to the following
equation.
CuSO4(aq) + Na2CO3(aq) CuCO3(s) + Na2SO4(aq)
The mass of copper (II) carbonate formed when 200cm3 of a solution
containing 5.3g of sodium carbonate per liter of solution was reacted
completely with excess copper (II) sulphate is given by
A. 5.3 𝑥 200 𝑥 124
𝑔
5.2 𝑥 124 𝑥 1000 106 𝑥 200 𝑥 124 106 𝑥 124 𝑥 100
B. 106 𝑥 200 𝑔 C. 5.3 𝑥 1000 𝑔 D. 5.3 𝑥 200 𝑔
106 𝑥 1000
27 Lead (II) nitrate reacts with potassium iodide according to the following
equation
Pb(NO3)2 (aq) + 2KI (aq) PbI2(s) + 2KNO3(aq)
The mass pf lead (II) iodide formed when 33.2g of potassium iodide is reacted
with excess lead (II) nitrate is (K= 39, I= 127, Pb = 207)
A. 4.61g B. 9.22g C. 46.1g D. 92.2g
29 Which one of the following solutions contains the same number of moles of
sodium ions as 200cm3 of o.5M NaHSO4 solution?
A. 100cm3 of 2M Na2CO3
B. 100cm3 of 0.5M NaNO3
C. 250cm3 of 0.8M NaHCO3
D. 250cm3 0.4M NaCl
30 10cm3 of monobasic acid completely reacted with 20cm3 of 0.05M sodium
carbonate solution. The number of moles of the acid that reacted is
20 𝑥 0.05 𝑥 2
A. � 1000 � 𝑚𝑜𝑙𝑒𝑠
20 𝑥 0.05 𝑥 2
B. � 10
� 𝑚𝑜𝑙𝑒𝑠
20 𝑥 0.05
C. � 2 𝑥1000 � 𝑚𝑜𝑙𝑒𝑠
0.05 𝑥 2 𝑥 10
D. � 20 𝑥 1000
� 𝑚𝑜𝑙𝑒𝑠
31 Iron react with oxygen to form 0.8g of Iron (III) oxide is [O = 16, Fe = 56]
0.8 𝑥 2 𝑥 56
A. � 160 � 𝑔
0.8 𝑥 2 𝑥 56
B. � 320
�𝑔
0.8 𝑥 2
C. �160 𝑥 56� 𝑔
0.8 𝑥 56
D. � 320 𝑥 2 � 𝑔
32 Nitric acid reacts with copper (II oxide according to the following equation
CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l)
0.5g of an impure copper (II) oxide reacted completely with 50cm3 of a 0.1M
nitric acid. The mass of copper (II) oxide in a sample is
A. 0.20g B. 0.24g C. 0.30g D. 0.40g
Moles of HCl
From equation 1 mole of Na2CO3 reacts with 2 moles of HCl
Mole of HCl = 2 x Moles of Na2CO3 = 0.002 x 2 = 0.004 moles
Volume of HCl solution
0.25moles are contained in 1000cm3 of .25M HCl solution
0.004 𝑥 1000
∴ 0.004 moles are in 0.25
= 16𝑐𝑚3
13 B Hint: Same number of moles of an element contain the same number of atoms
𝑚𝑎𝑠𝑠
Mole = 𝑟𝑎𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠
8
Mole of sulphur in 8g = 32 = 0.25moles
20
Moles calcium in 20g =40 = 0.5 moles
10
Moles calcium in 10g = 4 = 0.25 mole
Therefore, 8g of sulphur contain the same number of atoms as 10g of calcium
14 B Formula mass of H2SO4 = 1 x 2 + 32 + 16 x 4 = 98g
Mass of H2SO4 in 1000cm3 of 0.2M solution = (98 x 0.2)g
0.2 𝑥 98 𝑥 5
∴ 5cm3 contain 1000 g
Moles of HCl
From equation 1 mole of Na2CO3 reacts with 2 moles of HCl
Mole of HCl = 2 x Moles of Na2CO3 = (0.00125 x 2) moles
Molarity
22.7cm3 contain 0.00125 x 2 mole
0.00125 𝑥 2 𝑥 1000
∴ 1000cm3 contain 22.7
M
16 C Mole of acid =
25 𝑥 0.25
= 0.00625 moles
1000
25 𝑥 0.5
Moles sodium hydroxide = 1000
= 0.0125 𝑚𝑜𝑙𝑒𝑠
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑑𝑖𝑢𝑚 ℎ𝑦𝑑𝑟𝑜𝑥𝑖𝑑𝑒 0.0125
Basicity of acid = = =2
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑐𝑖𝑑 0.00625
17 D Formula mass of CaCO3 = 40 + 12 + 16 x 3 = 100g
Formula mass of CO2 = 12 + 16 x 2 = 44g
100g of CaCO3 produce 44 g of CO2
44 𝑥 20
20 g of CaCO3 produce 100 𝑔 𝑜𝑓 𝐶𝑂2
18 D (2 x 64) g of Cu produce 2(64 + 16) g of CuO
0.64 𝑥 2 𝑥 80 0.64 𝑥 80
0.64g of Cu produce 2 𝑥 64 = 64
19 B Formula mass of KI = 39 + 127 = 166
Formula mass PbI2 = 207 + 127 x 2 =461
166 x 2g of KI produce 461g of PbI2
461 𝑥 33.2
33.2g of KI produce 166 𝑥 2 = 46.1g of PbI2
Concentration in gl-1
1mole weigh = 36.5g
∴ 0.125 moles of HCl weigh 0.125 x 36.5 =4.5625
Therefore, concentration of HCl gl-1 = 4.5625
37 (a) Moles of NaOH
1000cm3 contain 0.4 moles
0.4 𝑥 30
30cm3 contain 1000 = 0.012 moles
38 Moles of NaOH
1000cm3 contain 0.2 moles
0.2 𝑥 25
25cm3 contain 1000 = 0.005 moles
Reaction equation
2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)
Moles of H2SO4
2moles of NaOH 1 mole of H2SO4
0.005 𝑥 1
0.005𝑚𝑜𝑙𝑒𝑠 of NaOH react with 2
= 0.0025 𝑚𝑜𝑙𝑒𝑠
3
24.6cm contain 0.0025 moles
0.0025 𝑥 1000
1000cm3 contain 24.60
= 0.1𝑀
39 Moles of HCl
1000cm3 contain 0.8 moles
0.8 𝑥 25
25cm3 contain 1000 = 0.02 moles
Reaction equation
2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g)
Moles of Na2CO3
2moles of HCl 1 mole of Na2CO3
0.02 𝑥 1
0.02𝑚𝑜𝑙𝑒𝑠 of HCl react with 2 = 0.01 𝑚𝑜𝑙𝑒𝑠
20cm3 contain 0.01 moles
0.01 𝑥 1000
1000cm3 contain 20
= 0.5𝑀
Formula mass of Na2CO3 = 2 x 23 + 12 + 16 x 3 = 106
mole of Na2CO3 weigh 106g
0.5moles weigh 106 x 0.5 = 53g
∴ the concentration of Na2CO3 is 53gl-1