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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

25

d

20

answer is C.

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Detailed Solution

milli-equivalent of NH₃. reacted with HNO₃
=45 x 0.4-20x0.1=16
$\begin{array}{l} ∴ \frac{W}{17} \times 1000 = 16; W_{{NH}_{3}} = 0 .272 g \\ wt . of N = 0 .272 \times \frac{14}{17} = 0 .224 \end{array}$
% N in the sample $= \frac{0.224}{1.12} \times 100 = 20 %$

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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: