0.115 g of a dibasic acid required 25 mL of 0.1 N NaOH solution for complete neutralization. Find the molecular weight of the acid. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
80
B
47
C
56
D
92

Solution:

For complete neutralization, number of equivalent of NaOH is equal to number of equivalent of dibasic acid.

$\begin{matrix} No . of equivalents of NaOH & = & No . of equivalents of dibasic acid \\ Normality \times Volume (in L) & = & \frac{Mass}{\frac{Molecular weight}{2}} \\ 0.1 N \times 25 mL \times \frac{1 L}{1000 mL} & = & \frac{0.115 g \times 2}{Molecular weight} \\ Molecular weight & = & 92 \end{matrix}$

$0.115 g$ of a dibasic acid required $25 mL$ of $0.1 N NaOH$ solution for complete neutralization. Find the molecular weight of the acid.

Solution:

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