13. In a laboratory a student prepared 0.2mN * a_{2} * S * O_{4} solution and another one prepared 0.2M N*a_{2} * S * O_{4} solution for a particular experiment. How these solution differ in their concentration?
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n-factor far Na
2
CO
3
=2
∵ The n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by one mole of the salt. And there is the exchange of 2 electrons so its n factor is 2
Therefore, Normality = n-fator × molarity
Molarity =
2
0.2
=0.1 M
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