Eq wt / n-factor

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 ?

5 Answers

106
Asish Mahapatra ·

in 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

11
Tush Watts ·

Acc to me, n - factor of A4 B3 shuld be = 12

23
qwerty ·

ya 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 ?

11
Tush Watts ·

Considering A4B3 to be a normal salt, the n - factor of a salt is the total charge of the anion.

1
ujjwalkalra kalra ·

hi 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.

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