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