Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol^-1]

0.300g

0.250g

0.2242g

0.448g

Correct answer is D

M = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)

= \(\frac{MmIT}{96500n}\)

Copper II Chloride = CuCl₂

CuCl₂ → Cu²⁺ + 2Cl²

Mass of compound deposited = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol^-1

no of charge = + 2

Mass = \(\frac{64 \times 1350}{96500 \times 2}\)

Mass = 0.448g

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Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol^-1]

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Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol^-1]