What is the time (in seconds) required to deposit 0.405 g of calcium from molten CaCl₂ using a current of 1.5 A? (Atomic
Ca²⁺ + 2e⁻ → Ca . 1 mol Ca (40.5 g) requires 2F. Moles = (0.405/40.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C . t = (Q/I) = (1930/1.5) = 1286.67 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement