How much solid oxalic acid (Molecular weight | vedclass

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(C) The normality $(N)$ of the solution is given by the formula: $N = \frac{	ext{Number of equivalents}}{	ext{Volume in Liters}}$.Number of equivale

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$0.63$

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(C) The normality $(N)$ of the solution is given by the formula: $N = \frac{	ext{Number of equivalents}}{	ext{Volume in Liters}}$.Number of equivalents $(n_{eq})$ $= N 	imes V(L) = 0.1 	imes \frac{100}{1000} = 0.01$.Oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ has a molecular weight of $126$. Its n-factor is $2$ (since it provides $2 \ H^+$ ions).Equivalent mass $= \frac{	ext{Molecular weight}}{n	ext{-factor}} = \frac{126}{2} = 63$.Weight required $= n_{eq} 	imes 	ext{Equivalent mass} = 0.01 	imes 63 = 0.63 \ g$.

How much solid oxalic acid (Molecular weight $126$) has to be weighed to prepare $100 \ mL$ of exactly $0.1 \ N$ oxalic acid solution in water (in $g$)?

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