11Considering A4B3 to be a normal salt, the n - factor of a salt is the total charge of the anion.
I have a small doubt.
suppose in a reaction of A4 B3 ,
A gains 3 e- and B loses 2 e-
then what will be the n - factor of A4 B3 in that reaction ? i.e the ratio of Mol wt and equivalent wt ? and hence wat will be its euqivalent weight ?
can i say that the n -factor will the NET e- lost or gained by A4 B3
hence 4 atoms of A will gain 12 e- total , and B loses total 6 e-, so is the n factor = 12-6 = 6 ??
and suppose it were a case such that total e- gained by A = total e- lost by B .. then ?
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5 Answers
106in almost all the cases, (that i have seen), no. of electrons gained by A = that lost by B..
i havent seen such a case yet (but as u know im not good @ chem) .
so im not sure
23ya i also havet seen such a case, but i thought aisa hua to ?? LoL i m worst in chem [3] [3]
ok.. so if total e- gained by A = total e- lost by B = 12
then n factor = 12 ... tushar u mean this only na ?
111hi this is the sure solution of ur problem...
if in reaction such case exsist then for total take the difference....and if both gain then intead of difference take the sum.