Question

In Kjeldahl's method of estimation of nitrogen, the ammonia evolved from 0.2 g of an organic compound was absorbed in 60 mL of 0.1 M $H_2SO_4$. If 40 mL of 0.1 M NaOH is required for complete neutralisation of the unused acid, the percentage of nitrogen in the compound is:

• A. 14
• B. 28
• C. 56
• D. 42

Hint:

For high nitrogen content, ensure calculation and conversion accuracy, particularly in converting mass to moles and understanding molar ratios in reactions.

Answer 1

The Correct Option is C

In Kjeldahl's method, nitrogen in an organic compound is converted into ammonia (\(NH_3\)). This ammonia is absorbed in a known volume of standard sulfuric acid (\(H_2SO_4\)). The unreacted sulfuric acid is then titrated against a standard solution of sodium hydroxide (\(NaOH\)). 
Given data:

• Weight of organic compound = 0.2 g
• Volume of \(H_2SO_4\) = 60 mL
• Molarity of \(H_2SO_4\) = 0.1 M
• Volume of \(NaOH\) = 40 mL
• Molarity of \(NaOH\) = 0.1 M

Step 1: Calculate the milliequivalents of initial \(H_2SO_4\): \[ \text{Normality of } H_2SO_4 = \text{Molarity} \times \text{Basicity} = 0.1 \times 2 = 0.2 \, N \] \[ \text{Milliequivalents of initial } H_2SO_4 = \text{Normality} \times \text{Volume (in mL)} = 0.2 \times 60 = 12 \, \text{meq} \] Step 2: Calculate the milliequivalents of \(NaOH\) used for titration: \[ \text{Normality of } NaOH = \text{Molarity} \times \text{Acidity} = 0.1 \times 1 = 0.1 \, N \] \[ \text{Milliequivalents of } NaOH = \text{Normality} \times \text{Volume (in mL)} = 0.1 \times 40 = 4 \, \text{meq} \] Step 3: Calculate the milliequivalents of unused \(H_2SO_4\): \[ \text{Milliequivalents of unused } H_2SO_4 = 4 \, \text{meq} \] Step 4: Calculate the milliequivalents of \(H_2SO_4\) reacted with \(NH_3\): \[ \text{Milliequivalents of reacted } H_2SO_4 = \text{Milliequivalents of initial } H_2SO_4 - \text{Milliequivalents of unused } H_2SO_4 \] \[ = 12 - 4 = 8 \, \text{meq} \] Step 5: Milliequivalents of \(NH_3\) and Nitrogen: \[ \text{Milliequivalents of } NH_3 = 8 \, \text{meq} \] \[ \text{Milliequivalents of Nitrogen} = 8 \, \text{meq} \] Equivalent weight of Nitrogen = 14 g/equivalent. 
Step 6: Calculate the weight of Nitrogen: \[ \text{Weight of Nitrogen} = \frac{\text{Milliequivalents of Nitrogen} \times \text{Equivalent weight of Nitrogen}}{1000} \] \[ = \frac{8 \times 14}{1000} = \frac{112}{1000} = 0.112 \, g \] Step 7: Calculate the percentage of Nitrogen in the organic compound: \[ \text{Percentage of Nitrogen} = \frac{\text{Weight of Nitrogen}}{\text{Weight of Organic compound}} \times 100 \] \[ = \frac{0.112}{0.2} \times 100 = 0.56 \times 100 = 56\% \] Correct Answer: (3) 56