The document defines key concepts regarding moles, molar mass, and concentration calculations: - A mole is the amount of substance containing 6.02x10^23 elementary units like atoms or molecules, and refers to the number of atoms/molecules in exactly 12 grams of carbon-12. - Molar mass is the mass of one mole of a substance in grams and allows conversion between mass and moles using the formula mass (g) = moles x molar mass (g/mol). - Concentration is the amount of solute per volume of solution, commonly measured in molarity (M) which is moles of solute per liter of solution. Dilution decreases concentration by adding more

1. MOLE CONCEPT
 Mole is an amount of substance which contain as many elementary
particles as there are in 12 grams of carbon-12.
 They contain the same number of particles in 12g of carbon-12.
 The number of individual particles in one mole of substance is called
AVOGADRO’S CONSTANT.
 Avogadro’s constant (L) = 6.02x𝟏𝟎 𝟐𝟑
particles.
 Number of atoms (N) = amount of substance(n)× L.
 The S.I unit of amount of substance is mole (mol)
 1 mol = 6.02x𝟏𝟎 𝟐𝟑

2.  Mathematically, Amount of substance (n) =
𝑚𝑎𝑠𝑠(𝑚)
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
 Molar mass (M) is the mass of 1 mole of a substance. Its unit is g/mol.
 From the equation: m = n×M and M =
𝑚
𝑛
.

3. WORKED EXAMPLE
 Calculate the number of moles in 5.85g of NaCl. [ Na = 23, Cl = 35.5]
solution
Number of moles(n) =
𝒎𝒂𝒔𝒔(𝒎)
𝑴𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔(𝑴)
where
m = 5.85g and M = 23+35.5 = 58.5g/mol
Hence: n =
𝟓.𝟖𝟓𝒈
𝟓𝟖.𝟓𝒈/𝒎𝒐𝒍
= 0.1 mol.
 Magnesium ribbon of mass 4.0g is placed in dilute hydrochloric acid (HCl)
contained in a beaker. Calculate the number of moles of HCl that would be
required to react completely with the ribbon. [Mg = 24, H = 1.0, Cl =35.5 ].

4. SOLUTION
 From the reaction:
𝑀𝑔(𝑠) + 2𝐻𝐶𝑙(𝑎𝑞) → 𝑀𝑔𝐶𝑙2 (𝑎𝑞) + 𝐻2 (𝑔)
From the balanced chemical equation:
n (Mg) =
𝑚
𝑀
=
4.0
24
= 0.167 mol
If 1 mol of Mg = 2 mol of 0.167mol.
Then 0.167mol is :
2
1
× 0.167 = 0.334 𝑚𝑜𝑙
Hence 0.334 moles of HCl react completely with the ribbon.

5. PREPARATION OF SOLUTION OF A GIVEN CONCENTRATION
 A solution is a uniform mixture of two or more substances. It contains a
solute and a solvent.
 Concentration is the amount of dissolved substances in a given volume of
solvent.
 Generally, Concentration, C =
𝐴𝑚𝑜𝑢𝑛𝑡 𝑂𝑓 𝑆𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑛
𝑉𝑜𝑙𝑢𝑚𝑒 (𝑉)
 The units of concentration is mol/𝑑𝑚3 or M.
M is Molarity which expresses the number of moles of a solute per litre of a
solution.

6. Worked Example
 An aqueous solution of volume 2.0𝑑𝑚3 contains 25.5g NaCl. Calculate the
concentration of the solution. [ NaCl = 58.5 ]
solution
Concentration, C =
𝑛
𝑉
, where
v = 2.0𝑑𝑚3
and n =
𝑚
𝑀
=
25.5
58.5
= 0.436 mol
Hence C =
0.436𝑚𝑜𝑙
2.0𝑑𝑚3 = 0,218 mol/𝑑𝑚3

7. TRY:
 In a laboratory experiment, a student was required to prepare 500𝑐𝑚3 of
1.0 M solution of glucose (𝐶6 𝐻12 𝑂6 ). Determine the
(i) molar mass of glucose.
(ii) Amount of substance in moles in the solution,
(iii) mass of glucose contained in the solution.
[ C = 12 , H = 1.0 , O = 16 ]
Answer
(i) 180g/mol
(ii) 0.5 mol
(iii) 90g

8. DILUTION
 Dilution is the process of reducing the concentration of a solute in a solution by
addition of more solvent. A concentrated solution from which a dilute solution is
prepared is called stock solution.
 In diluting a solution;
Amount of substance in concentrated solution 𝑛1 = Amount of substance in dilute
solution 𝑛2 .
If n = C × 𝑣
then 𝐶1 𝑉1 = 𝐶2 𝑉2
Where 𝐶1 = Initial concentration
𝐶2 = Final concentration
𝑉1 = Initial volume
𝑉2 = Final volume

9. Worked Example
 A solution of 2.0M HCl is diluted to 500𝑐𝑚3 of 0.1M HCl. Determine the
initial volume of the acid.
solution
𝐶1 𝑉1 = 𝐶2 𝑉2
where 𝐶1 = 2.0M ; 𝐶2 = 0.1M ; 𝑉2 = 500𝑐𝑚3
Hence 𝑉1 =
𝐶2 𝑉2
𝐶1
=
0.1×500
2.0
= 25𝑐𝑚3
Initial volume, 𝑉1 = 25𝑐𝑚3.

10. TRY:
 Determine the volume of water required to change the concentration of
100.0𝑐𝑚3
of 0.5M HCl to 0.1M HCl.
ans: 500.0𝑐𝑚3
 If 500.0𝑐𝑚3 of 1.0M NaOH solution is diluted to 1.0𝑑𝑚3 with distilled
water, calculate the concentration of the diluted solution.
ans: 0.5𝑑𝑚3