Question

The number of electrons present in 100 ml of 0.1 n h2so4 is

Answer 1

∵ normality = n × molarity

In case of H₂SO₄ ⇔2H⁺ + SO₄²⁻ ,
n - factor = 2
so, molarity = normality/2 = 0.1N/2 = 0.05M
∴ 0.05 M = number of mole of sulphuric acid /volume of solution in L
⇒0.05 × 0.1 = 0.005 = number of mole of sulphuric acid

Now, number of electron in a molecule of H₂SO₄ = 2 + 16 + 4 × 8
= 2 + 16 + 32 = 50
∴ number of electrons in 0.005 mole = 50 × 0.005 × 6.023 × 10²³
= 250 × 6.023 × 10²⁰
= 1505.75 × 10²⁰ = 1.50575 × 10²³

Answer 2

.........................................