One mole of a substance is the amount containing as many elementary entities as the number of atoms in exactly 12 grams of carbon-12. It is denoted with n.

1. $$\text{number of moles} = \frac{m}{M}$$
2. $$\text{number of moles} = \frac{CV}{1000}$$
3. $$\text{number of moles= CV}$$
4. $$n = \frac{PV}{RT}$$
5. $$n = \frac{\text{Number of particles}}{N_A}$$
where m = mass in gram
M = molar mass in g/mol
P = pressure in atm
C = Concentration in mol/dm³
V = Volume in cm³ or dm³
R = molar gas constant
T = Temperature in kelvin
N_A= 6.02 × 10²³
Note: n = CV when V is in dm³
n = CV/1000 when V is in cm³

Example 1: Find the number of moles present in 20g of CaCO₃

$$\text{number of moles} = \frac{m}{M}$$ $$\text{M of }CaCO_3= 40 + 12 + 48$$ $$\text{M} = 100g/mol$$ $$\text{number of moles} = \frac{20}{100}$$ $$\text{number of moles} = 0.2 \text{ mol}$$

Example 2: What is the concentration of a solution containing 2g of NaOH in 100Cm³ of solution? [Na = 23, H =1, O = 16](Jamb)

Example 3: What is the molar mass of a substance, if 0.4 mole of the Substance has a mass of 25.0g (Jamb)

Example 4: How many atoms are present in 6.0g of Magnesium? [Mg = 24, N_A = 6.02 × 10²³] (JAMB)

$$\text {Mass} = {6g} $$ $$\text {Molar mass} = {24} $$ $${N_A} = {6.02} × {10^{23}} $$ $$\text{number of moles} = \frac{mass}{molar mass} $$ $$\text{n} = \frac{6}{24} $$ $$\text{n} = {0.25} $$ $$\text{n} = \frac{\text{number of particles}}{N_A} $$ $$\text{number of atoms} = {n} × {N_A} $$ $$ = {0.25} × {6.02 × 10^{23}} $$ $$\text{number of atoms} = {1.51} × {10^{23}} $$

Example 5: What is the number of oxygen atoms in 32g of the gas? (WAEC)

$$\text{Oxygen gas is diatomic} = {O_2} $$ $$\text{molar mass of } {O_2} = {16 × 2} $$ $$ = 32g/mol $$ $$\text{mass of oxygen} = {32g} $$ $$\text{number of atoms} = {n} × {N_A} $$ $$ = \frac{mass}{\text{molar mass}} × {N_A} $$ $$ = \frac{32}{32} × {6.02} × {10^{23}} $$ $$\text{number of atoms} = {6.02} × {10^{23}} $$

Example 6: Find the concentration of a solution prepared by dissolving 15g of KOH in H₂O to make 2L of solution

$$\text{Mass of KOH} = {15g} $$ $$\text{Volume} = {1L} = {1dm^3 } $$ $$ \text{molar mass of KOH} = {39 + 16 +1} $$ $$ = {56}\text{ g/mol} $$ $$ {n} = \frac{15} {56} $$ $$ {n} = {0.268} \text{ moles} $$ $$ {n} = \text{concentration × volume} $$ $$\text{concentration} = \frac{n}{V} $$ $$\text{concentration} = \frac{0.268}{2} $$ $$\text{concentration} = {0.134}\text{ mol/dm³} $$

Example 7: How many moles of AgNO₃ are there in 500cm³ of 0.10mol/dm³ AgNO₃ solution (WAEC)

$$ {Concentration} = {0.10} \text{ mol/dm³} $$ $$ {volume} = {500cm³} $$ $$\text{since volume is in cm³} $$ $$ {n} = \frac{CV}{1000} $$ $$ {n} = \frac{ 0.1 × 500}{1000} $$ $$ \text{number of moles} = {0.005} \text{ moles} $$

Example 8: Pure sulphuric acid is a liquid of 1.84 density. What volume of it would be required to prepare 250cm³ of a 0.2mol/dm³ solution? (JAMB)

$${density} = {1.84}\hspace{0.2em}{kgm^{-3} } $$ $$ {volume} = {250cm³} $$ $$ {concentration} = {0.2} \text{moldm³} $$ $$\text{molar mass of }{H_2SO_4}\text { ,M} $$ $$ = {1 × 2 + 32 + 16 × 4} = {98}\text{ g/mol} $$ $$ {n} = \frac{0.2 × 250}{1000} $$ $$ {n} = \text{0.05 moles} $$
Volume at its density can be calculated using; $$ {density} = \frac{mass} {volume} $$ $${but,} \text{ mass} = {n × \text{Molar mass}} $$ $$ {mass} = {0.05 × 98} = {4.9g} $$ $$ \text{volume at density} = \frac{mass}{Volume} $$ $$ \text{volume at density} = \frac{4.9}{1.84} $$ $$ \text{volume at density} = {2.66}\text{ dm³} $$