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  4. 0.14 gm of an acid required 12.5 ml of 0.1 N NaOH for complete neturalisation. The equivalent mass of the acid is:

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Q. $0.14$ gm of an acid required $12.5\, ml$ of $0.1 N \,NaOH$ for complete neturalisation. The equivalent mass of the acid is:

Organic Chemistry – Some Basic Principles and Techniques

Solution:

$mEq$. of $NaOH =12.5 \times 0.1$ $=1.25$
mEq. of acid $=$ mEq. of $NaOH$
Equivalent $=\frac{ Wt }{ Ew }$
$\therefore EW =\frac{ Wt .}{\text { Equivalent }}=\frac{0.14}{1.25 \times 10^{-3}}$ $=112$