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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

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detailed solution

Correct option is A

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

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

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

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

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

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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