(i) Measurement of atomic mass of an element is somewhat more complicated since most of the elements exist as a mixture of isotopes, each of which has its own mass.
(ii) Thus, it is essential to consider this isotopic mixture while calculating atomic mass of an element.

Atomic mass unit is one - twelfth of mass of carbon - 12 atom, an isotope of carbon which contains 6 protons and 6 neutrons.

(i) One mole (6.023 x 10²³ of entities) of any gas occupies 22.4 litre or 22400 ml at S.T.P.
(ii) This volume is called as molar volume of gas.

Hydrogen chloride (HCl), Hydrogen iodide (HI) are two examples for heterodiatomic molecules.

The total number of atoms present in the molecule is called its atomicity.
\(\text { Atomicity }=\frac{\text { Molecular mass }}{\text { Atomic mass }}\)

1)  0.3 mole of aluminium (Atomic mass of Al = 27)
\(\text { Number of moles }=\frac{\text { Mass of } \mathrm{Al}}{\text { Atomic mass of } \mathrm{Al}}\)
Mass = No. of moles × atomic mass
So, mass of Al = 0.3 × 27
= 8.1 g
2) 2.24 litre of SO₂ gas at S.T.P
 Molecular mass of SO₂ = 32 + (16 × 2)
= 32 + 32 = 64
\(\text { Number of moles of } \mathrm{SO}_2=\frac{\begin{array}{c} \text { Given volume of } \mathrm{SO}_2 \text { at S.T.P } \end{array}}{\begin{array}{c} \text { Molar volume } \mathrm{SO}_2 \text { at S.T.P } \end{array}}\)
Number of moles of SO₂ = \(\frac{2.24}{22.4}\)
= 0.1 mole
\(\text { Number of moles }=\frac{\text { Mass }}{\text { Molecular mass }}\)
Mass = No. of moles × molecular mass
Mass = 0.1 × 64
Mass of SO₂ = 6.4 g
iii) 1.51 × 10²³ molecules of water
Molecular mass of H₂O = 18
\(\text { Number of moles }=\frac{\begin{array}{c} \text { Number of molecules of }\text { water } \end{array}}{\text { Avogadro's number }}\)
= 1.51 × 10²³ / 6.023 × 10²³
= 1 / 4
= 0.25 mole
\(\text { Number of moles }=\frac{\text { Mass }}{\text { Molecular mass }}\)
0.25 = mass / 18
Mass = 0.25 × 18
Mass = 4.5 g
iv) 5 × 10²³ molecules of glucose
Molecular mass of glucose = 180
\(\text { Mass of glucose }=\frac{\begin{array}{c} \text { Molecular mass } \times \text { number of particles } \end{array}}{\text { Avogadro's number }}\)
= (180 × 5 × 10²³) / 6.023 × 10²³
= 149.43 g

Molecular mass of water (H₂O) = H₂O = H x 2 + O x 1
= 1 x 2 + 16 x 1
= 2 + 16
=18
No of moles = \(\frac{Given \ Mass}{Molecular \ Mass}\)
Number of moles = \(\frac{0.18}{8}= \frac{0.18 \times 100}{18 \times 100}=0.01\)
No of moles = \(\frac{Number\ of \ molecules}{Avogadro's \ number}\)
Number of molecules = No of moles x Avogadro's number
= 0.01 x 6.023 x 10²³
= 0.06023 x 10²³
Number of molecules in 0.18 g of water = 6.023 x 10²¹
Alternative method
Gram molecular mass of water = H₂O
= 1 x 2 + 16 x 1 = 2 + 16
= 18 g
Number of molecules \(=\frac{\text { Avogadro's number } \times \text { given mass }}{\text { Gram molecular mass }}\)
Number of molecules \(=\frac{6.023 \times 10^{23} \times 0.18}{18}\) 
= \(6.023 \times 10^{23} \times 0.01\)
Number of molecules of 0.18 g of water = 6.023 x 10²¹ molecules.

The atomicity of an element can be derived using Avogadro's hypothesis. Let us consider the following equation

According to Avogadro's law, 1 volume of any gas occupy 'n' number of molecules.

if n = 1 then

(i) 1 molecule of hydrogen chloride gas is made up of \(\frac{1}{2}\) molecule of hydrogen and \(\frac{1}{2}\) molecule of chlorine
(ii) But 1 molecule of hydrogen chloride contains one atom of hydrogen and I atom of chlorine
(iii) Hence \(\frac{1}{2}\) molecule of hydrogen = 1 atom of hydrogen
(iv) Or 1 molecule of hydrogen = 2 atoms of hydrogen
(v) So the atomicity of hydrogen is 2, and molecular formula is H₂
Similarly,
\(\frac{1}{2}\) molecule of chlorine = 1 atom of chlorine 1molecule of chlorine = 2 atoms of chlorine So the atomicity of chlorine is 2, and its molecular formula is Cl₂

Applications of Avogadro's hypothesis
(i) It explains Gay-Lussac's law.
(ii) It helps in the determination of atomicity of gases.
(iii) Molecular formula of gases can be derived using Avogadro's law.
(iv) It determines the relation between molecular mass and vapour density.
(v) It helps to determine Gram molar volume of all gases (i.e. 22.4 lit at S.T.P)

We consider B-11 isotope presents x % in nature. So B-10, isotope presents (1-x) %.
Average atomic mass = mass of B- 11+ mass of B-10
10.804 = x \(\times\) 11 + (1 - x) \(\times\) 10 
10.804 = 11 x + 10 -10 x 
10.804 = x + 10 
x = (10.804 - 10) 
x = 0.804
So, % relative abundance of B-11 is 0.804 x 10 = 80.4 %
% relative abundance of B -10 is (1 - x) x 100
=(1 - 0.804) x 100 
= 0.196 x 100 
= 19.6 %
% relative abundance of B-10 and B-11 are 19.6 % and 80.4 % respectively.

\(\text { Molar mass of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}=\mathrm{Al} \times 2+3[\mathrm{~S} \times 1+\mathrm{O} \times 4]\)
\(27 \times 2+3[32 \times 1+16 \times 4]\)
= 54 + 3[32 + 64]
= 54 + 3 [96]
Molar Mass = 54 + 288 = 342
\(\text { Mass } \% \text { of an element }=\frac{\text { mass of that element in the compound }}{\text { molar mass of the compound }} \times100\)
\(\text { Mass } \% \text { oxygen in } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}=\frac{192}{342} \times 100 \)
\(=0.5614 \times 100=56.14 \% \)
\(\text { Mass } \% \text { of oxygen in } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \text { is } 56.14 \%\)

Number of moles = \(\frac{Mass}{Atomic \ mass}\)
∴ Mass = Number of moles x Atomic mass
= 0.5 x 16
= 8g.