Question

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

Answer 1

Answer:

normality = n × molarity. In case of H₂SO₄ ⇔2H⁺ + SO₄²⁻ , n - factor = 2 so, molarity = normality/2 = 0.1N/2 = 0.05 M

Answer 2

Explanation:

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