Question

When 1000 C of charge is passed through CuSO4 solution, x g of copper is deposited. How much charge should be passed through the solution to deposit 5x g of copper?​

Answer 1

Number of charge = 5018 C

Further explanation

Given

1000 C of charge for x grams of copper

Required

Number of charge

Solution

Faraday's Law :

$\tt W=\dfrac{e.i.t}{96500}\\\\W=\dfrac{e.Q}{96500}$

For 1000 C, W = x grams

$\tt x=\dfrac{1000.e}{96500}=0.0104e$

For 5x grams :

$\tt 5x=\dfrac{e.Q}{96500}\\\\5\times 0.0104e=\dfrac{e.Q}{96500}\\\\Q=\dfrac{96500\times 5\times 0.0104e}{e}=5018~C$