Question:
In the estimation of nitrogen by Kjeldahl's method, the ammonia evolved from 0.30 g of an organic compound (X) was passed into 100 mL of 0.1 M $\text{H}_2\text{SO}_4$. The unreacted acid required 20 mL of 0.5 M NaOH for complete neutralization. What is X?
In the estimation of nitrogen by Kjeldahl's method, the ammonia evolved from 0.30 g of an organic compound (X) was passed into 100 mL of 0.1 M $\text{H}_2\text{SO}_4$. The unreacted acid required 20 mL of 0.5 M NaOH for complete neutralization. What is X?
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Remember that Urea contains two nitrogen atoms per molecule, making its nitrogen content exceptionally high (46.7%).
This value is a standard constant in organic biochemistry questions.
This value is a standard constant in organic biochemistry questions.
Updated On: Jul 22, 2026
- $\text{CH}_3\text{CONH}_2$
- $\text{C}_6\text{H}_5\text{CONH}_2$
- $(\text{NH}_2)_2\text{CO}$
- $\text{C}_6\text{H}_5\text{NH}_2$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
This is a quantitative analysis problem based on Kjeldahl's method for nitrogen estimation.
We need to calculate the mass percentage of nitrogen in the organic compound X and identify the compound from the options.
Step 2: Key Formula or Approach:
The percentage of nitrogen ($\% \text{N}$) in the sample of mass $W$ can be determined using:
\[ \% \text{N} = \frac{1.4 \times N \times V}{W} \] where:
$N$ is the normality of the acid used to neutralize ammonia.
$V$ is the volume of the acid (in mL) neutralized by ammonia.
$W$ is the mass of the organic compound (in grams).
Step 3: Detailed Explanation:
• Let us calculate the initial milliequivalents (meq) of $\text{H}_2\text{SO}_4$:
For $\text{H}_2\text{SO}_4$, the basicity is 2, so Normality ($N$) = $2 \times \text{Molarity} = 2 \times 0.1 = 0.2\text{ N}$.
Initial meq of $\text{H}_2\text{SO}_4$ = $\text{Normality} \times \text{Volume in mL} = 0.2\text{ N} \times 100\text{ mL} = 20\text{ meq}$.
• Let us calculate the milliequivalents of unreacted $\text{H}_2\text{SO}_4$ neutralized by NaOH:
For NaOH, the acidity is 1, so Normality ($N$) = $1 \times \text{Molarity} = 0.5\text{ N}$.
Meq of NaOH used = $0.5\text{ N} \times 20\text{ mL} = 10\text{ meq}$.
Since meq of acid neutralized = meq of base, the unreacted acid = $10\text{ meq}$.
• Let us find the milliequivalents of acid reacted with the evolved ammonia ($\text{NH}_3$):
Meq of acid reacted with $\text{NH}_3$ = $\text{Initial meq} - \text{Unreacted meq} = 20 - 10 = 10\text{ meq}$.
• Let us calculate the mass percentage of nitrogen:
Using the standard Kjeldahl formula with $N \times V = 10$:
\[ \% \text{N} = \frac{1.4 \times (N \times V)}{W} \] \[ \% \text{N} = \frac{1.4 \times 10}{0.30} = \frac{14}{0.30} = 46.67\% \]
• Now, we match this percentage with the given options:
(A) Acetamide ($\text{CH}_3\text{CONH}_2$, Molar mass = $59\text{ g/mol}$): $\% \text{N} = \frac{14}{59} \times 100 \approx 23.7\%$
(B) Benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$, Molar mass = $121\text{ g/mol}$): $\% \text{N} = \frac{14}{121} \times 100 \approx 11.6\%$
(C) Urea ($(\text{NH}_2)_2\text{CO}$, Molar mass = $60\text{ g/mol}$): $\% \text{N} = \frac{28}{60} \times 100 = 46.67\%$
(D) Aniline ($\text{C}_6\text{H}_5\text{NH}_2$, Molar mass = $93\text{ g/mol}$): $\% \text{N} = \frac{14}{93} \times 100 \approx 15.0\%$
The nitrogen percentage of 46.67% matches perfectly with Urea.
Step 4: Final Answer:
The organic compound X is urea, $(\text{NH}_2)_2\text{CO}$.
This is a quantitative analysis problem based on Kjeldahl's method for nitrogen estimation.
We need to calculate the mass percentage of nitrogen in the organic compound X and identify the compound from the options.
Step 2: Key Formula or Approach:
The percentage of nitrogen ($\% \text{N}$) in the sample of mass $W$ can be determined using:
\[ \% \text{N} = \frac{1.4 \times N \times V}{W} \] where:
$N$ is the normality of the acid used to neutralize ammonia.
$V$ is the volume of the acid (in mL) neutralized by ammonia.
$W$ is the mass of the organic compound (in grams).
Step 3: Detailed Explanation:
• Let us calculate the initial milliequivalents (meq) of $\text{H}_2\text{SO}_4$:
For $\text{H}_2\text{SO}_4$, the basicity is 2, so Normality ($N$) = $2 \times \text{Molarity} = 2 \times 0.1 = 0.2\text{ N}$.
Initial meq of $\text{H}_2\text{SO}_4$ = $\text{Normality} \times \text{Volume in mL} = 0.2\text{ N} \times 100\text{ mL} = 20\text{ meq}$.
• Let us calculate the milliequivalents of unreacted $\text{H}_2\text{SO}_4$ neutralized by NaOH:
For NaOH, the acidity is 1, so Normality ($N$) = $1 \times \text{Molarity} = 0.5\text{ N}$.
Meq of NaOH used = $0.5\text{ N} \times 20\text{ mL} = 10\text{ meq}$.
Since meq of acid neutralized = meq of base, the unreacted acid = $10\text{ meq}$.
• Let us find the milliequivalents of acid reacted with the evolved ammonia ($\text{NH}_3$):
Meq of acid reacted with $\text{NH}_3$ = $\text{Initial meq} - \text{Unreacted meq} = 20 - 10 = 10\text{ meq}$.
• Let us calculate the mass percentage of nitrogen:
Using the standard Kjeldahl formula with $N \times V = 10$:
\[ \% \text{N} = \frac{1.4 \times (N \times V)}{W} \] \[ \% \text{N} = \frac{1.4 \times 10}{0.30} = \frac{14}{0.30} = 46.67\% \]
• Now, we match this percentage with the given options:
(A) Acetamide ($\text{CH}_3\text{CONH}_2$, Molar mass = $59\text{ g/mol}$): $\% \text{N} = \frac{14}{59} \times 100 \approx 23.7\%$
(B) Benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$, Molar mass = $121\text{ g/mol}$): $\% \text{N} = \frac{14}{121} \times 100 \approx 11.6\%$
(C) Urea ($(\text{NH}_2)_2\text{CO}$, Molar mass = $60\text{ g/mol}$): $\% \text{N} = \frac{28}{60} \times 100 = 46.67\%$
(D) Aniline ($\text{C}_6\text{H}_5\text{NH}_2$, Molar mass = $93\text{ g/mol}$): $\% \text{N} = \frac{14}{93} \times 100 \approx 15.0\%$
The nitrogen percentage of 46.67% matches perfectly with Urea.
Step 4: Final Answer:
The organic compound X is urea, $(\text{NH}_2)_2\text{CO}$.
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