$1.26 g$ of a dibasic acid were dissolved in water and made up to $200 m L . 20 m L$ of this solution were completely neutralised by $10 m L$ of $N / 5$ caustic soda solution. Calculate the equivalent mass and molecular mass of the acid.

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Solution

As we know that,
Normality $= \frac{m a s s o f s o l u t e}{e q u i v a l e n t m a s s} \times v o l u m e (L)$
Equivalent mass $= \frac{m o l e c u l a r m a s s}{n - f a c t o r}$
[ Here $n$ factor is 2 as it is dibasic acid ]
Normality of dibasic acid $= \frac{1.26}{x} \times \frac{1000}{200}$
$N_{1} = \frac{6.3}{x} \Rightarrow (x = e q . m a s s)$
By the formula,
$N_{1} V_{1} = N_{2} V_{2}$
$\frac{6.3}{x} \times 20 = 10 \times \frac{1}{5} \frac{6.3}{2} \times 20 = x \Rightarrow x = 68 E q u i v a l e n t m a s s = 63 a n d m o l e c u l a r m a s s = 63 \times 2 = 126 (2 i s n - f a c t o r)$

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1.26 g of a dibasic acid were dissolved in water and made up to 200 mL. 20 mL of this solution were completely neutralised by 10 mL of N/5 caustic soda solution. Calculate the equivalent mass and molecular mass of the acid.

$1.26 g$ of a dibasic acid were dissolved in water and made up to $200 m L . 20 m L$ of this solution were completely neutralised by $10 m L$ of $N / 5$ caustic soda solution. Calculate the equivalent mass and molecular mass of the acid.