0.50g of an organic compound was kjeldahlised and the $N H_{3}$ evolved was absorbed in a certain volume of 0.5M $H_{2} S O_{4}$ . The residual acid required 60 $c m^{3}$ of M/2 $N a O H$ . If the percentage of N is 56, then the volume of 0.5M $H_{2} S O_{4}$ taken was:

30 ml

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40 ml

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50 ml

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60 ml

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Solution

The correct option is C 50 ml
from kjeldahlised method:
% $N_{2} = \frac{1.4 \times m e q . o f H_{2} s o_{4} u s e d t o n e u t r a l i z e N H_{3}}{m a s s o f c o r p d}$
$= \frac{1.4 \times 2 M (v - \frac{v_{1}}{2})}{m}$
Val of $H_{2} s o_{4} o f M a l a r i t y \neq M [o r (2 M) n a r m a l i t y] = V m l .$
$V a l . o f N a o H o f M o l a r i t y M (o r (M) n o r m a l l y) = V_{1} m l .$
used for titration cocess of $H_{2} s o_{4}$
m.q of excers $H_{2} s o_{4} = m e r . q N a o H = M V_{1} m e r .$
Total $m e q . o f H_{2} s o_{4} t a k e n = 2 M V$
$m e q o f H_{2} s o_{4} u s e d = 2 M V - M V_{1} = m e q o f N H_{3}$
$∴ 56 = \frac{1.4 \times \times 2 \times 0.5 (v - \frac{60}{2})}{0.50} = 28 (V - 30) \Rightarrow 20 = v - 30$
$V = 50 m l .$

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