The molarity of \(0.2 \mathrm{~N} \mathrm{Na}_2 \mathrm{CO}_3\) solution will be
The molarity of \(0.2 \mathrm{~N} \mathrm{Na}_2 \mathrm{CO}_3\) solution will be
\(0.05 \mathrm{M}\)
\(0.2 \mathrm{M}\)
\(0.1 \mathrm{M}\)
\(0.4 \mathrm{M}\)
Solution
\(n\) - factor for \(\mathrm{Na}_2 \mathrm{CO}_3=2\)
\(\therefore\) 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 \(=n\)-factor \(\times\) molarity
\(\Rightarrow\) Molarity \(=0.2 / 2=0.1 \mathrm{M}\)
Hence, the correct option is (c).