What is the molarity of 0.4\ N\ Na2CO3 solution? a) 0.1 M b) 0.2 M c) 0.4 M d) 0.8 M

For $\mathrm{Na}{}_2\mathrm{CO}{}_3$, the normality $N$ is related to molarity $M$ by the number of replaceable hydrogen ions (or equivalent factor) per mole of solute. $\mathrm{Na}{}_2\mathrm{CO}{}_3$ is a dibasic salt and can release 2 $\mathrm{OH}{}^{-}$ ions per formula unit in a reaction with acid:

$$\mathrm{Na}{}_2\mathrm{CO}{}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+\mathrm{H}{}_2\mathrm{O}+\mathrm{CO}{}_2$$

Thus, the equivalent factor $n$ is 2.

Given: $$N=0.4$$

The relationship between normality and molarity is: $$N=n\times M$$

Rearranging to solve for $M$: $$M=\frac{N}{n}=\frac{0.4}{2}=0.2$$

The molarity of the solution is 0.2 M. The correct answer is (b) 0.2 M.