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From the following data calculate the va...

From the following data calculate the value of Avogadro constant. Charge of an electron `= 1.6 xx 10^(-19)` coulomb. (1 faraday = 96500 coulombs)

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To calculate the value of Avogadro's constant (N_A) using the given data, we can follow these steps: ### Step 1: Understand the relationship between charge, moles, and Avogadro's constant The charge of one mole of electrons (which is equivalent to 1 Faraday) can be expressed as: \[ \text{Charge on 1 mole of electrons} = N_A \times \text{Charge of 1 electron} \] ### Step 2: Substitute the known values From the question, we know: ...
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RC MUKHERJEE-ELECTROLYSIS AND ELECTROLYTIC CONDUCTANCE-Objective Problems
  1. From the following data calculate the value of Avogadro constant. Char...

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  2. The number of electrons involved in the reaction when one faraday of e...

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  3. Number of electrons involved in the electrodeposited of 63.5 g of Cu...

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  4. Faraday's laws of electrolysis are related to the

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  5. The electric of electricity produced m kg of a subtances X. Electroche...

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  6. 1 coulomb of electricity is passed through a solution of AlCl(3), 13.5...

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  7. Electrochemical equivalent of a substance is 0.0006735, its eq. wt. is

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  8. When electricity is passed through a solution of AlCl(3) and 13.5g of ...

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  9. 3.17 g of a substance was deposited by the flow of 0.1 mole of electro...

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  10. A current of 0.5 ampere when passed through AgNO(3) solution for 193 s...

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  11. During electrolysis of aqueous solution of a salt pH in the space near...

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  12. In the electrolysis of CuCl(2) solution (aq) with Cu electrodes, the w...

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  13. The current required to diplace 0.1 g of H(2) in 10 seconds will be

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  14. The charge of an electron is 1.6 xx 10^(-19) coulomb. How many electro...

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  15. 96500 coulombs deposit 107.9 g of Ag from its soluton. If e = 1.6 xx 1...

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  16. A current of 2.0 ampere is passed for 5.0 hour through a molten tin sa...

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  17. The cost at 5 paise/KWH of operating an electric motor for 8 hours whi...

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  18. One faraday of charge was passed through the electrolytic cells placed...

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  19. The time required (approx) to remove electrolytically one half from 0....

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  20. In the electrolysis of H(2)SO(4), 9.72 litres and 2.35 litres of H(2) ...

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  21. In the electrolysis of H(2)O, 11.2 litres of H(2) was liberated at cat...

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