Question

29.5 mg of organic compound contains N. Ammonia from Kjeldahl method absorbed in 20 mL of 0.1 M HCl. Excess acid requires 15 mL of 0.1 M NaOH for neutralization. The percentage of nitrogen in compound is

• A. 29.5%
• B. 59.0%
• C. 23.7%
• D. 47.4%

Hint:

Nitrogen percentage in Kjeldahl = \( \frac{14\times\text{mmol reacted}}{\text{sample mass(g)}}\times100 \).

Answer 1

The Correct Option is C

Step 1: Acid initially = 20 mL of 0.1 M \[ \text{milli-mole HCl} = 20 \times 0.1 = 2. \]
Step 2: Back-titrated by 15 mL NaOH \[ = 15 \times 0.1 = 1.5 \text{ milli-mole}. \]
Step 3: Acid reacted with NH$_3$ \[ = 2 - 1.5 = 0.5 \text{ milli-mole}. \]
Step 4: In Kjeldahl 1 mole N → 1 mole NH$_3$ neutralizes 1 mole HCl. Therefore \[ \text{mole of N in sample} = 0.5 \times 10^{-3}. \]
Step 5: Mass of N \[ = 14 \times 0.5 \times 10^{-3} = 7 \times 10^{-3} \text{ g}. \]
Step 6: Sample mass 29.5 mg = 0.0295 g. \[ %\;N = \frac{0.007}{0.0295} \times 100 = 23.7%. \] Hence → (C).