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.

i) \(\text { Number of moles of } \mathrm{CO}_2=\frac{\text { Volume at S.T.P }}{\text { Molar volume }}\)
= 11.2 / 22.4
= 0.5 mole
Number of molecules of CO₂ = number of moles of CO₂ × Avogadro’s number
= 0.5 × 6.023 × 10²³
= 3.011 × 10²³ molecules of CO₂
ii) \(\begin{aligned} & \text { Number of atoms } \text { of } \mathrm{Au}=\frac{\begin{array}{c} \text { Mass of Au } \times \text { Avogadro's } \text { number } \end{array}}{\text { Atomic mass of } \mathrm{Au}} \\ \end{aligned}\)
Number of atoms of Au =\(\cfrac { 1 }{ 198 } \times 6.023\times { 10 }^{ 23 }\)
Number of atoms of Au = 3.042 × 10²¹ g
iii) \(\text { Number of molecules }=\frac{\begin{array}{c} \text { (Avogadro number } \times \text { Given mass) } \end{array}}{\begin{array}{c} \text { Gram molecular } \text { mass } \end{array}}\)
Number of molecules of water = 6.023 × 10²³ × 54 / 18
= 18.069 × 10²³ molecules
iv) (a) 1 mole of CO₂ contains 2 moles of oxygen
(b) 5 moles of CO₂ contain 10 moles of oxygen
Number of atoms of oxygen = number of moles of oxygen × Avogadro’s number
= 10 × 6.023 × 10²³
= 6.023 × 10²⁴ atoms of Oxygen
(a)  1 mole of CO₂ contains 1 mole of carbon
(b) 5 moles of CO₂ contains 5 moles of carbon
No. of atoms of carbon = No. of moles of carbon × Avogadro’s number
= 5 × 6.023 × 10²³
= 3.011 × 10²⁴ atoms of Carbon

1) \(\text { Number of moles }=\frac{\text { Mass of the element }}{\begin{array}{c} \text { Atomic mass of the element } \\ \end{array}}\)
= 46/23
= 2 moles of sodium
2) 5.6 litre of oxygen at S.T.P
 \(\text { Number moles }=\frac{\text { Given volume of } \mathrm{O}_2 \text { at S.T.P }}{\text { Molar volume at S.T.P }}\)
Number of moles of oxygen = \(\cfrac { 5.6 }{ 22.4 } \)
= 0.25 mole of oxygen
iii) \(\text { Number of moles }=\frac{\text { Number of atoms of iron }}{\text { Avogadro's number }}\)
= 12.046 × 10²³ / 6.023 × 10²³
= 2 moles of iron

molecular mass of H₂SO₄
= (1 × 2) + (32 × 1) + (16 × 4)
= 2 + 32 + 64
= 98 g
% of S in H₂SO₄ = \(\frac { Mass\ of\ sulphur }{ Moleculer\ mass\ of\ H_2\ SO_4 } \times 100\)
% of S in H₂SO₄ = \(\cfrac { 32 }{ 98 } \times 100\)
= 32.65%