Question

How many electrons will be lost at anode by the anion when 5.85 g fused NaCl is electrolysed? ​

Answer 1

firstly we have to find moles of nacl

and here molecular weight of nacl is 58.5

so moles=wt/mwt

n=0.1

number of electron lost at anode is equal to

n*Na=N

N=6.022*10^22

Answer 2

Answer:

first you have to find moles of NACL and here molecular weight of NACL is 58.5

so moles = wt/mat

n = 0.1

number of electron lost at anode is equal to

n*Na = N

N = 6.022*10^22

Explanation:

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