Question

How many grams of water are present in 100ml of 0.1 m solution of Sulphuric acid?

Answer 1

Explanation:

n 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

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