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Q.

The NH3 evolved due to complete conversion of N from 1.12 g sample of protein was absorbed in 45 ml of 0.4 N HNO3. The excess acid required 20 mL of 0.1 N NaOH. The % N in the sample is:

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a

8

b

16

c

20

d

25

answer is C.

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Detailed Solution

milli-equivalent of NH3. reacted with HNO3=45 x 0.4-20x0.1=16∴W17×1000=16; WNH3=0.272gwt. of N=0.272×1417=0.224% N in the sample =0.2241.12×100=20%
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