i) 1 mole of calcium carbonate are involved
ii) Gram molecular mass of \(\mathrm{CaCO}_{3} \rightarrow \mathrm{Ca} \times 1+\mathrm{C} \times 1+\mathrm{O} \times 3\)
40 x 1 + 12 x 1 + 16 x 3 
40 + 12 + 48 = 100 g
Gram molecular mass of CaCO₃ is 100 g / mol.
iii) 1 mole of CO₂ are there in above equation.

1) \(\begin{array}{r} \text { Number of moles of } \mathrm{CO}_2=\frac{\text { Given volume at S.T.P }}{\text { Molar volume at S.T.P }} \\ \end{array}\)
\(\begin{array}{r} 2.5 \text { mole of } \mathrm{CO}_2=\frac{\text { Volume of } \mathrm{CO}_2 \text { at S.T.P }}{22.4} \end{array}\)
Volume of CO₂ at S.T.P = 22.4 × 2.5
= 56 litres.
2) 12.046 × 10²³ of ammonia gas molecules
\(\text { Number of moles }=\frac{\text { Number of molecules }}{\text { Avogadro's number }}\)
= 12.046 × 10²³ / 6.023 × 10²³
= 2 moles
Volume occupied by NH₃ =  number of moles × molar volume
= 2 × 22.4
= 44.8 litres at S.T.P
3) Number of moles = 14 / 28
= 0.5 mole
Volume occupied by N₂ at S.T.P
= no. of moles × molar volume
= 0.5 × 22.4
= 11.2 litres.

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

i) H₂O
Atomic masses of H = 1, O = 16
Gram molecular mass of H₂O
= (1 × 2) + (16 × 1)
= 2 + 16
Gram molecular mass of H₂O = 18 g
ii) CO₂
Atomic masses of C = 12, O = 16
Gram molecular mass of CO₂
= (12 × 1) + (16 × 2)
= 12 + 32
Gram molecular mass of CO₂ = 44 g
iii) Ca₃ (PO₄)₂
Atomic masses of Ca = 40, P = 30, O = 16.
Gram molecular mass of Ca₃ (PO₄)₂
= (40 × 3) + [30 + (16 × 4)] × 2
= 120 + (94 × 2)
= 120 + 188
Gram molecular mass of Ca₃(PO₄)₂ = 308 g

Molar mass of CH₄ = 12 + 4
= 16 g
Mass % of carbon = \(\cfrac { 12 }{ 16 } \) x 100
= 75 %
Mass % of hydrogen= = \(\cfrac { 4 }{ 16 } \)x 100
= 25 %