Explanation

(a)
Molarity = normality $\times \frac{equivalent\ weight\ }{molecular\ weight\ }$
Given, normality of $Na_{2}{CO}_{2}\$solution =0.2 N
Equivalent weight = M
Molecular weight 2 M ($\because{Na}_{2}{CO}_{3}\ is\ \ dipositive.$)
$\therefore$ Molarity $= \ 0.2\ \times \ \frac{M}{2M}$
$= 0.1\ M$