Question

Na2S2O3 has the molar mass M. In the reaction Na2S2O3 + H2O + Cl2 → Na2SO4 + 2HCl + S the equivalent weight of Na2S2O3 is :-

Answer 1

Answer:3m/8

Explanation:

Answer 2

Answer:

From the equation we get, Na2S2O3 + H2O + Cl2 → Na2SO4 + 2HCl + S.

Here, oxidation number of Na2S2O3 =+2.

And the oxidation number of Na2SO4 =+4.

Now, from the equation Na2S2O3 has 1 mole and sulphur is of 1 then oxidation number of Na2S2O3 is 1×2=2

In the product side Na2SO4 has 1 mole and sulphur has 1 mole then the oxidation number is 1×4=4

The difference between the exchanged of electron is 4-2=2.

Now, we know that the equivalent weight of Na2S2O3 is = molecular weight of Na2S2O3/n-factor.

Here, the N factor is the number of exchanged electron /number of moles.

Then the equivalent weight of Na2S2O3=158×2/2=158.