MOLE : QUESTION & ANSWER

Q1
Calculate the mass of 6.022 × 10²³ molecule of Calcium carbonate (CaCO₃).

Solution —
Molar mass (Molecular mass in gram) of CaCO₃ = 100 g
No. of moles of CaCO3
= No. of molecules/Avogadro constant
= 6.022 × 10²³/ 6.022 × 10²³
= 1 mole
Mass of CaCO₃ 
= No. of moles × molar mass
= 1 × 100 g =  100 g.

Q2
Calculate the mass of 12.044 × 10²³ carbon atoms.

Solution —
No. of moles of Carbon atoms
= No. of atoms/Avogadro constant
= 12.044 × 10²³/6.022 × 10²³
= 2 mole
Mass of carbon atoms
= No. of moles × atomic mass
= 2 × 12
= 24 g.

Q3
Calculate the number of oxygen atoms in 1 mole of O₂.

Solution —
1 molecule of O₂ = 2 oxygen atoms
So, 1 mole of O₂ = 2 mole oxygen atoms
= 2 × 6.022 × 10²³ = 12.044 ×10²³ oxygen atoms.

Q4
Calculate the number of moles, and number if atoms of H, S, and O in 5 mole of H₂SO₄.

Solution —
1 mole of H₂SO₄ contains 2 mole of H, 1 mole of S, and 4 mole of O
=> 5 mole of H₂SO₄ contains
10 mole of H = 10 × 6.022×10²³ = 6.022×10²⁴ H atoms
5 mole of S = 5 × 6.022×10²³ = 3.011×10²⁴ S atoms
20 mole of O = 20 × 6.022×10²³ = 1.204×10²⁵ O atoms.

Q5
Calculate the number of oxygen atoms and its mass in 50 g of CaCO₃.

Solution —
Molecular mass of CaCO₃= 40+12+3×16 = 100

No. mole of CaCO₃ = 50g/100g = 0.5

0.5 mole of CaCO₃ contains 1.5 moles of oxygen atoms

No. of oxygen atoms = 1.5 × 6.022×10²³ = 9.033×10²³ atoms

Mass of Oxygen atoms = 1.5 × 16 = 24 g.

Q6
Calculate the number of atoms of each element in 122.5 g of KClO₃.

Solution —
Molecular mass of KClO₃ = 39 +35.5+3×16 = 122.5
No. of mole of KClO₃ = 122.5g/122.5g = 1 mole

1 mole of KClO₃ contains
1 mole of K = 6.022×10²³ K atoms
1 mole of Cl = 6.022×10²³ Cl atoms
3 mole of O = 3 × 6.022×10²³ = 1.806 ×10²⁴ O atoms.