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Q.a) Write Faraday's first law of electrolysis. b) The conductivity of 0.01 M KCl solution at 298 K is 0.00141 S cm^-1. Calculate the molar conductivity of 0.01 M KCl solution. OR

a) Write Faraday's second law of electrolysis. b) The conductivity of 0.05 M NaOH solution at 298 K is 0.01150 S cm^-1. Calculate the molar conductivity of 0.05 M NaOH solution.
Rajasthan RbseRajasthan Board Senior Secondary Examination 2025Subjective· 3mImportance★★★★★
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Faraday's first law relates the mass deposited at an electrode to the charge passed; using Lambda_m = kappa x 1000/M with the given kappa = 0.00141 S/cm and M = 0.01 mol/L gives a molar conductivity of 141 S cm^2 mol^-1.

a) FARADAY'S FIRST LAW OF ELECTROLYSIS: the mass (m) of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (charge, Q) passed through the electrolyte:

m proportional to Q

m = Z x Q = Z x i x t

where Z is the electrochemical equivalent of the substance, i is the current, and t is the time for which it flows.

b) MOLAR CONDUCTIVITY CALCULATION:

Given: kappa (conductivity) = 0.00141 S cm^-1, Molarity M = 0.01 mol/L

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