Question

The molarity of $0.2\, N\, Na _{2} CO _{3}$ solution will be

• A. 0.05 M
• B. 0.2 M
• C. 0.1 M
• D. 0.4 M

Answer 1

Answer: (C)

$n$-factor far $Na _{2} CO _{3}=2$
$\because$ 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 $\times$ molarity
Molarity $=\frac{0.2}{2}=0.1\, M$