Relative molecular mass:
Relative molecular mass of a gas or vapour is the ratio between the mass of one molecule of gas or vapour to mass of one atom of hydrogen.
Relative molecular mass \(=\frac{\text { Mass of } 1 \text { molecule of a gas (or) vapour at STP }}{\text { Mass of } 1 \text { atom of hydrogen }}\)     .......(1)
Vapour density:
It is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
Vapour density (V.D) \(=\frac{\text { Mass of a given volume of gas(or) vapour at S.T.P }}{\text { Mass of the same volume of the hydrogen }}\)
According to Avogadro's law [Let the number of molecules in one volume = 'n']
Vapour density (V.D) \(=\frac{\text { Mass of 'n' molecules of a gas (or) vapour at S.T.P }}{\text { Mass of ' } n \text { 'molecules of a hydrogen }}\)
cancelling 'n' vapour density (V.D) \(=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P }}{\text { Mass of } 1 \text { molecule of a hydrogen }}\)
Hydrogen is diatomic molecule, so
Vapour density \(=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P }}{\text { Mass of } 2 \text { atoms of hydrogen }}\)
Vapour density \(=\frac{\text { mass of } 1 \text { molecule of a gas or vapour at S.T.P }}{2 \times \text { mass of 1atom of hydrogen }}\)
2 x Vapour density \(=\frac{\text { mass of } 1 \text { molecule of a gas (or) vapour at S.T.P }}{\text { mass of 1atom of hydrogen }}\)
From eqn (1)
2 x Vapour density = Relative molecular mass

(i) 27 g of AI
Number of moles = \(\frac{Given\ Mass}{Atomic \ Mass}=\frac{27}{27}=1\)
Number of moles of 27 g of Al is 1 mole
(ii) 1.51 x 10²³ molecules of NH₄Cl
Number of moles = \(\frac{Number \ of \ molecules}{6.023 \times 10^\text{23}}\)
= \(\frac{1.51 \times10^{23}}{6.023 \times10^{23}}\) 
\(\frac{1.51}{6.023}\)
Number of moles of 1.51 x 10²³ molecules of NH₄Cl = 0.25 mole

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.

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.

\(\text { Molar mass of calcium carbonate } \mathrm{CaCO}_{3}=\mathrm{Ca} \times 1+\mathrm{C} \times 1+\mathrm{O} \times 3 \)
\(=40 \times 1+12 \times 1+16 \times 3\) 
\(=40+12+18=100 \mathrm{~g}\)
\(\text { Mass } \% \text { of an element }=\frac{\text { mass of that element in the compound }}{\text { molar mass of the compound }} \times 100  \)
\(\text { Mass } \% \text { of calcium }=\frac{40}{100} \times 100=40 \%  \)
\(\text { Mass } \% \text { of carbon }=\frac{12}{100} \times 100=12 \%  \)
\(\text { Mass } \% \text { of oxygen }=\frac{48}{100} \times 100=48 \%\)