(C) The $n$-factor for $Na_2CO_3$ is $2$ because it acts as a base and can accept $2 \ H^+$ ions.Normality is related to molarity by the formula: $\te

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(C) The $n$-factor for $Na_2CO_3$ is $2$ because it acts as a base and can accept $2 \ H^+$ ions.Normality is related to molarity by the formula: $\text{Normality} = n\text{-factor} \times \text{Molarity}$.Given: $\text{Normality} = 0.2 \ N$ and $n\text{-factor} = 2$.Therefore,$\text{Molarity} = \frac{\text{Normality}}{n\text{-factor}} = \frac{0.2}{2} = 0.1 \ M$.Hence,the correct option is $C$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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The molarity of $0.2 \ N$ $Na_2CO_3$ solution will be (in $M$)

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A
$0.05$
B
$0.2$
C
$0.1$
D
$0.4$

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Some Basic Concepts of Chemistry

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The molarity of $0.2 \ N$ $Na_2CO_3$ solution will be (in $M$)

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