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:

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%

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:

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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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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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 $= \frac{0.224}{1.12} \times 100 = 20 %$