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  4. 100 mL of 0.1 N NaOH solution is required for neutralisation of 0.49 g of a dibasic acid. What is the molecular weight of the acid?

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Q. $100\, mL$ of $0.1 \,N\,NaOH$ solution is required for neutralisation of $0.49\, g$ of a dibasic acid. What is the molecular weight of the acid?

Some Basic Concepts of Chemistry

Solution:

Milliequivalents of acid = Milliequivalents of base

$\frac{0.49}{E} \times 1000=0.1 \times 100$ (Say equivalent wt. of acid $=E$ )

$\therefore E=49$

Molecular weight = Equivalent weight $\times$ Basicity $=49 \times 2 =98$