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

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(C) For neutralization,the number of equivalents of acid equals the number of equivalents of base.$N_1 V_1 = N_2 V_2$Given: Weight of acid $(W)$ = $0.

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(C) For neutralization,the number of equivalents of acid equals the number of equivalents of base.$N_1 V_1 = N_2 V_2$Given: Weight of acid $(W)$ = $0.16 \ g$,Volume of $NaOH$ $(V_2)$ = $25 \ mL$,Normality of $NaOH$ $(N_2)$ = $0.1 \ N$ (decinormal).Using the formula: $\frac{W}{E} = N_2 	imes \frac{V_2}{1000}$$\frac{0.16}{E} = 0.1 	imes \frac{25}{1000} = \frac{2.5}{1000} = 0.0025$$E = \frac{0.16}{0.0025} = 64$Since the acid is dibasic,the molecular weight $(M)$ is $2 	imes E$.$M = 2 	imes 64 = 128$.

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

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