25 , mL of a decinormal NaOH solution is comp | vedclass

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(C) For neutralization,the number of equivalents of acid must equal the number of equivalents of base.Equivalents of $NaOH = N 	imes V(L) = \frac{1}{

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(C) For neutralization,the number of equivalents of acid must equal the number of equivalents of base.Equivalents of $NaOH = N 	imes V(L) = \frac{1}{10} 	imes \frac{25}{1000} = 0.0025 \, eq$.Let the molecular weight of the dibasic acid be $M$.The equivalent weight of the dibasic acid is $E = \frac{M}{2}$ (since it is dibasic).Equivalents of acid = $\frac{	ext{mass}}{E} = \frac{0.16}{M/2} = \frac{0.32}{M}$.Equating the two: $\frac{0.32}{M} = 0.0025$.$M = \frac{0.32}{0.0025} = 128$.

$25 \, mL$ of a decinormal $NaOH$ solution is completely neutralized by $0.16 \, g$ of a dibasic acid. The molecular weight of the dibasic acid is:

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