In Kjeldahl's method for the estimation of nitrogen, $$1.4\text{ g}$$ of an organic compound was digested with concentrated $$H_2SO_4$$ and the mixture was then distilled with an excess of $$NaOH$$. The ammonia gas evolved was completely absorbed in $$50\text{ mL}$$ of $$0.5\text{ M } H_2SO_4$$.

The residual unreacted acid required $$60\text{ mL}$$ of $$0.5\text{ M } NaOH$$ for complete neutralization.

Calculate the percentage by mass of nitrogen in the organic compound.

(Round off to the nearest integer)

We need to calculate the percentage by mass of nitrogen in the organic compound using the data from Kjeldahl's method.

Step-by-Step Calculation:

1. Total Initial Moles of $$\text{H}_2\text{SO}_4$$:

   $$\text{Initial millimoles of }\text{H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (mL)} = 0.5\text{ M} \times 50\text{ mL} = 25\text{ mmol}$$



2. Moles of Unreacted $$\text{H}_2\text{SO}_4$$ (from neutralization with $$\text{NaOH}$$):

   The neutralization reaction is:

   $$2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$ $$\text{Millimoles of }\text{NaOH} = 0.5\text{ M} \times 60\text{ mL} = 30\text{ mmol}$$ $$\text{Millimoles of unreacted }\text{H}_2\text{SO}_4 = \frac{1}{2} \times \text{Millimoles of }\text{NaOH} = \frac{30}{2} = 15\text{ mmol}$$



3. Moles of $$\text{H}_2\text{SO}_4$$ Consumed by Ammonia ($$\text{NH}_3$$):

   $$\text{Millimoles of }\text{H}_2\text{SO}_4\text{ reacted with }\text{NH}_3 = 25\text{ mmol} - 15\text{ mmol} = 10\text{ mmol}$$



4. Moles and Mass of Nitrogen Atoms:

   Since $$1\text{ mol}$$ of $$\text{H}_2\text{SO}_4$$ neutralizes $$2\text{ mol}$$ of $$\text{NH}_3$$ (each containing one nitrogen atom):

   $$\text{Millimoles of Nitrogen (N)} = 2 \times \text{Millimoles of consumed }\text{H}_2\text{SO}_4 = 2 \times 10\text{ mmol} = 20\text{ mmol}$$ $$\text{Mass of Nitrogen} = \text{Moles} \times \text{Molar Mass} = \frac{20}{1000}\text{ mol} \times 14\text{ g mol}^{-1} = 0.28\text{ g}$$



5. Percentage of Nitrogen in the Compound:

   $$\%\text{ Nitrogen} = \left( \frac{\text{Mass of Nitrogen}}{\text{Mass of Organic Compound}} \right) \times 100$$ $$\%\text{ Nitrogen} = \left( \frac{0.28\text{ g}}{1.4\text{ g}} \right) \times 100 = 20\%$$

Conclusion:

The stoichiometric analysis of the back-titration confirms that the organic sample contains exactly $$20\%$$ nitrogen by mass.

Answer: Option B — 20