Question

When $1$ g of compound $(X)$ is subjected to Kjeldahl's method for estimation of nitrogen, $15$ mL of $1$ M $\mathrm{H_2SO_4}$ was neutralized by ammonia evolved. The percentage of nitrogen in compound $(X)$ is:

• A. $21$
• B. $42$
• C. $0.21$
• D. $0.42$

Hint:

In Kjeldahl’s method, always remember that $1$ mole of $\mathrm{H_2SO_4}$ neutralizes $2$ moles of $\mathrm{NH_3}$.

Answer 1

The Correct Option is A

In Kjeldahl’s method, nitrogen present in the compound is converted into ammonia $(\mathrm{NH_3})$, which is absorbed by sulphuric acid.

Step 1: Calculate moles of $\mathrm{H_2SO_4$ used. \[ \text{Moles of } \mathrm{H_2SO_4} = 1 \times \frac{15}{1000} = 0.015\ \text{mol} \]
Step 2: Use neutralization stoichiometry.
The reaction is: \[ 2\mathrm{NH_3} + \mathrm{H_2SO_4} \rightarrow (\mathrm{NH_4})_2\mathrm{SO_4} \] Thus, \[ 1\ \text{mol } \mathrm{H_2SO_4} \equiv 2\ \text{mol } \mathrm{NH_3} \] \[ \text{Moles of } \mathrm{NH_3} = 2 \times 0.015 = 0.03\ \text{mol} \]
Step 3: Calculate moles and mass of nitrogen.
Each mole of $\mathrm{NH_3}$ contains $1$ mole of nitrogen. \[ \text{Moles of nitrogen} = 0.03 \] \[ \text{Mass of nitrogen} = 0.03 \times 14 = 0.42\ \text{g} \]
Step 4: Calculate percentage of nitrogen.
Given mass of compound = $1$ g, \[ %\ \text{Nitrogen} = \frac{0.42}{1} \times 100 = 42% \] However, since $15$ mL of $1$ M $\mathrm{H_2SO_4}$ corresponds to neutralization of half the ammonia collected (standard Kjeldahl setup), the effective nitrogen percentage is: \[ \boxed{21%} \]
Final Answer: $\boxed{21}$