0.16 g of a dibasic acid required 25 mL of | vedclass

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(C) For complete neutralization,the number of equivalents of the acid must equal the number of equivalents of the base.Equivalents of $NaOH = 	ext{No

Vedclass · Accepted Answer

$128$

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(C) For complete neutralization,the number of equivalents of the acid must equal the number of equivalents of the base.Equivalents of $NaOH = 	ext{Normality} 	imes 	ext{Volume (in L)} = \frac{1}{10} 	imes \frac{25}{1000} = 0.0025 \ eq$.Since the acid is dibasic,its equivalent weight is $\frac{M}{2}$,where $M$ is the molecular weight.Equivalents of acid = $\frac{	ext{Mass}}{	ext{Equivalent Weight}} = \frac{0.16}{M/2} = \frac{0.32}{M}$.Equating the two: $\frac{0.32}{M} = 0.0025$.$M = \frac{0.32}{0.0025} = 128 \ g/mol$.

$0.16 \ g$ of a dibasic acid required $25 \ mL$ of decinormal $NaOH$ solution for complete neutralization. The molecular weight of the acid is

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