Question

In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from $0.75\, g$ of sample neutralized $10\, mL$ of ${1 \, M \, H_2SO_4}$. The percentage of nitrogen in the soil is :

• A. 37.33
• B. 45.33
• C. 35.33
• D. 43.33

Answer 1

The Correct Option is A

To determine the percentage of nitrogen in the soil sample using Kjeldahl's method, we follow these steps:

1. Ammonia reacts with sulfuric acid to form ammonium sulfate. The reaction can be represented as follows: 2 \, NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4.
2. The sulfuric acid used is 10\, mL of 1\, M solution, which contains:
   • Moles of H_2SO_4 = \frac{1 \, \text{mol/L} \times 10 \, \text{mL}}{1000} = 0.01 \, \text{mol}
3. From the balanced chemical equation, 2 moles of NH_3 react with 1 mole of H_2SO_4. Therefore, moles of NH_3 are double the moles of H_2SO_4:
   • Moles of NH_3 = 2 \times 0.01 = 0.02 \, \text{mol}
4. The molar mass of nitrogen (N) is approximately 14 \, \text{g/mol}. Therefore, the mass of nitrogen in the evolved NH_3 is:
   • Mass of nitrogen = 0.02 \, \text{mol} \times 14 \, \text{g/mol} = 0.28 \, \text{g}
5. The percentage of nitrogen in the soil sample is calculated using:
   • \text{Percentage of Nitrogen} = \left(\frac{\text{Mass of nitrogen}}{\text{Total mass of sample}}\right) \times 100
   • Substituting the values, we get: \text{Percentage of Nitrogen} = \left(\frac{0.28 \, \text{g}}{0.75 \, \text{g}}\right) \times 100 = 37.33\%

Therefore, the correct answer is 37.33\%, which corresponds to the option 37.33. Thus, the percentage of nitrogen in the soil sample is 37.33\%.