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0.1914g of an organic acid is dissolved in approx. 20 ml of water. 25 ml of 0.12 N NaOH required for the complete neutralization of the acid solution. The equivalent weight of the acid is

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63.80

62.50

answer is C.

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Detailed Solution

$Normality = \frac{W}{GEW \times V_{(lit)}} \Rightarrow 0.12 = \frac{0.1914 \times 1000}{GEW \times 25}$

GEW $= \frac{0.1914 \times 1000}{0.12 \times 25} = 63.8 .$

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