A) 0.1 M
B) OM
C) 0.4 M
D) 0.2 M
Correct Answer: A
Solution :
\[\text{Molarity}\,\text{=}\,\text{normality}\,\text{=}\frac{\text{equivalen}\,\text{weight}}{\text{molecular}\,\text{weight}}\] Given, normality of \[\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\]solution \[\text{=}\,\text{0}\text{.2 N}\] Equivalent weight = M Molecular weight = 2 M (\[\because \,\,\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\]is dipositive.) \[\text{Molarity}=0.2\times \frac{M}{2M}\] \[=0.1\,\text{M}\]
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