Question

n factor of na2s2o3 how to calculate

Answer 1

n-factor is the no. of moles of e- lost or gained by substance during course of rxn..in this case Na2SO3 dissociate like
Na2SO3====>2Na+ + SO3 -2
here SO3 is gaining two moles of electorn hence n-factor will be 2

Answer 2

N factor of $Na_2 S_2 O_3$ is 1.

Solution:

N-factor is the no. of moles of e- lost or gained by substance during course of reaction. In this case $Na_2 S_2 O_3$ dissociate like,

$2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}+2 \mathrm{Nal}$

First calculate the oxidation state of Sulphur is reactant side, it is +2.

Then calculate the oxidation state in the product side, it is +2.5.

Change in oxidation state gives n factor which for this particular reaction is 0.5.

But, as there are 2 Sulphur atoms in Sodium thiosulphate, change in oxidation state is  $2 \times 0.5=1$.

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