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Q.

0.303g of organic compound was analyzed for nitrogen by Kjeldahl's method. The ammonia gas evolved was absorbed in 50ml of 0.05M $H_{2} {SO}_{4}$ . The excess acid required 25ml of 0.1M NaOH for neutralization. The percentage of nitrogen in the given compound is

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a

11.5

b

12.5

c

23

d

14.5

answer is A.

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Detailed Solution

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$% N = \frac{1.4 \times (N_{acid} V_{acid} - N_{NaOH} V_{NaOH})}{Wt . of organic compound}$

$= \frac{1.4 (50 \times 0.1 - 25 \times 0.1)}{0.303}$

= 11.5%

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