Question

the amount of silver deposited from a solution of silver nitrate when a current of 965 coulombs was passed is
who knows better answer and explain it only they can give answer ..​

Answer 1

hey ur ans is 1.08 gm.. ....

Answer 2

Given: charge, Q = 965 C

To Find: Amount of silver deposited, m.

Solution:

To calculate, the formula used:

• Q = number of electrons(n)x number of moles(N) x Faraday's constant
• Q = n x ( m/M) x F
• Here, n = 1

        Faraday's constant, F = 96500

        Number of moles, N = mass of silver deposited(m) /  Molar mass(M)

        Molar mass of silver M = 108g

Applying the above formula:

965 = 1 x (m/M )x 96500

965 = 1 x ( m/ 108) x 96500

965 / 96500 = m / 108

m/108 = 965/ 96500

m / 108 = 1 / 100

m = (1/100) x 108

   = 108 / 100

   = 1.08

m = 1.08 g

Hence, the amount of silver deposited is 1.08g.