The number of Hydrogen ions in 1.0$dm^{3}$ of 0.02$moldm^{-3}$ tetraoxosulphate(VI) acid is   $[N_{A} = 6.02 \times 10^{23}]$

Analyze the question: We need the number of H⁺ ions in 1.0 dm³ of 0.02 mol/dm³ H₂SO₄ (tetraoxosulphate(VI) acid).

Concept: H₂SO₄ is a strong dibasic acid. Each mole of H₂SO₄ gives 2 moles of H⁺ when it dissociates completely:

H₂SO₄ → 2H⁺ + SO₄²⁻

Number of particles = moles × Avogadro's number (N_A).

Solution:

1. Moles of H₂SO₄ = concentration × volume = 0.02 × 1.0 = 0.02 mol
2. Moles of H⁺ = 2 × 0.02 = 0.04 mol
3. Number of H⁺ ions = 0.04 × 6.02 × 10²³ = 0.2408 × 10²⁴ ≈ 2.4 × 10²²

Check: 0.04 × 6.02 × 10²³ = 0.2408 × 10²³ = 2.408 × 10²². This gives 2.4 × 10²².

(Be careful with the powers of ten: 0.04 = 4 × 10⁻², so 4 × 6.02 × 10²³ × 10⁻² = 24.08 × 10²¹ = 2.4 × 10²².)

Answer: The number of hydrogen ions is 2.4 × 10²², which is option C.

Tip: For acids, always check the basicity (number of replaceable H⁺). H₂SO₄ gives 2 H⁺, so multiply the moles of acid by 2. Also, convert decimals to standard form before multiplying so you don't slip on powers of ten. Option A (1.2 × 10²²) is the trap from forgetting the factor of 2.