0.45 g of an acid (mol wt. = 90) required 20 | vedclass

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(B) For complete neutralization,the number of equivalents of acid equals the number of equivalents of base.Equivalents of $KOH = N 	imes V 	ext{ (in

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$2$

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(B) For complete neutralization,the number of equivalents of acid equals the number of equivalents of base.Equivalents of $KOH = N 	imes V 	ext{ (in Liters)} = 0.5 	imes \frac{20}{1000} = 0.01 	ext{ equivalents}$.Equivalents of acid $= \frac{	ext{Mass}}{	ext{Equivalent weight}} = \frac{0.45}{	ext{Eq. Wt}} = 0.01$.$	ext{Eq. Wt} = \frac{0.45}{0.01} = 45$.$	ext{Basicity} = \frac{	ext{Molecular weight}}{	ext{Equivalent weight}} = \frac{90}{45} = 2$.

$0.45 \ g$ of an acid (mol wt. $= 90$) required $20 \ mL$ of $0.5 \ N$ $KOH$ for complete neutralization. The basicity of the acid is:

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