A sample of organic material weighing 1.058 g is decomposed by the Kjeldahl method. The ammonia is distilled into 50.0 mL of0.100 M HCl solution. The excess remaining acid required 10.10 mL of0.098 M NaOH solution. The percentage of nitrogen in the sample would be

# A sample of organic material weighing 1.058 g is decomposed by the Kjeldahl method. The ammonia is distilled into 50.0 mL of0.100 M $\mathrm{HCl}$ solution. The excess remaining acid required 10.10 mL of0.098 M $\mathrm{NaOH}$ solution. The percentage of nitrogen in the sample would be

1. A

5.45%

2. B

10.90%

3. C

12.4%

4. D

16.7%

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### Solution:

Amount of $\mathrm{HCl}$ to start with  $=\left(50.0×{10}^{-3}\mathrm{L}\right)\left(0.10\mathrm{molL}=50.0×{10}^{-4}\mathrm{mol}$

Amount of excess $\mathrm{HCl}$ $\mathrm{HCl}=\left(10.0×{10}^{-3}\mathrm{L}\right)\left(0.098\mathrm{molL}=9.8×{10}^{-4}\mathrm{mol}$

Amount of ${\mathrm{NH}}_{3}$ evolved or amount of N in the compound  $=\left(50.0-9.8\right)×{10}^{-4}\mathrm{mol}=41.2×{10}^{-4}\mathrm{mol}$

Mass of N in the compound $=\left(41.2×{10}^{-4}\mathrm{mol}\right)\left(14\mathrm{g}{\mathrm{mol}}^{-1}\right)=0.05768\mathrm{g}$

Percentage of N in the compound $=\frac{0.05768}{1.058}×100=5.45\mathrm{%}$  