0.63 g of oxalic acid is dissolved to obtain | vedclass

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Question

(C) Molar mass of oxalic acid $(COOH)_2 \cdot 2H_2O = 126 \ g/mol$. Equivalent mass of oxalic acid = $\frac{	ext{Molar mass}}{n	ext{-factor}} = \fra

Vedclass · Accepted Answer

$0.04$

Vedclass · Answer

(C) Molar mass of oxalic acid $(COOH)_2 \cdot 2H_2O = 126 \ g/mol$. Equivalent mass of oxalic acid = $\frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{126}{2} = 63 \ g/eq$. Number of gram equivalents = $\frac{	ext{Given mass}}{	ext{Equivalent mass}} = \frac{0.63}{63} = 0.01 \ eq$. Normality $(N) = \frac{	ext{Number of gram equivalents}}{	ext{Volume of solution in } L} = \frac{0.01}{250/1000} = \frac{0.01 	imes 1000}{250} = 0.04 \ N$. Therefore,the normality of the solution is $0.04 \ N$.

$0.63 \ g$ of oxalic acid is dissolved to obtain $250 \ cm^3$ of its solution. Find the normality of this solution. [Oxalic acid: $(COOH)_2 \cdot 2H_2O$] (in $N$)

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