Question

Ammonia evolved from \(0.75\,\text{g}\) of the soil sample in the Kjeldahl’s method for nitrogen estimation neutralises \(10\,\text{mL}\) of \(1\,\text{M}\) \(H_2SO_4\). Find the percentage of nitrogen present in the soil.

• A. \(35.33%\)
• B. \(37.33%\)
• C. \(43.33%\)
• D. \(45.33%\)

Hint:

In Kjeldahl’s method:

(1,textmol H2SO4 equiv 2,textmol NH3)
Always relate ammonia evolved to nitrogen using stoichiometry

Answer 1

The Correct Option is B

Step 1: Calculate the moles of sulphuric acid used. \[ \text{Volume of } H_2SO_4 = 10\,\text{mL} = 0.01\,\text{L} \] \[ \text{Molarity} = 1\,\text{M} \] \[ \text{Moles of } H_2SO_4 = 1 \times 0.01 = 0.01\,\text{mol} \] Step 2: Use the reaction between ammonia and sulphuric acid. \[ H_2SO_4 + 2NH_3 \rightarrow (NH_4)_2SO_4 \] From the equation: \[ 1 \text{ mol } H_2SO_4 \text{ reacts with } 2 \text{ mol } NH_3 \] \[ \text{Moles of } NH_3 = 2 \times 0.01 = 0.02\,\text{mol} \] Step 3: Calculate the mass of nitrogen present. Each mole of \(NH_3\) contains 1 mole of nitrogen. \[ \text{Moles of N} = 0.02 \] \[ \text{Mass of N} = 0.02 \times 14 = 0.28\,\text{g} \] Step 4: Calculate the percentage of nitrogen in the soil. \[ %\text{ of N} = \frac{0.28}{0.75} \times 100 = 37.33% \] Final Answer: \[ \boxed{37.33%} \]