How many coulombs are required to deposit 0.965 g of barium from a BaCl₂ solution? (Atomic mass of Ba = 137 g/mol, F = 9
Ba²⁺ + 2e⁻ → Ba . 1 mol Ba (137 g) requires 2F. Moles = (0.965/137) = 0.00704 mol , Charge = 0.00704 × 2 × 96500 = 1358.24 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry