Question:
A sample of $0.50 \,g$ of an organic compound was treated according to Kjeldahl?? method. The ammonia evolved was absorbed in $50 \,mL$ of $0.5\, M \,H_{2}SO_{4}$. The residual acid required $60\, mL$ of $0.5 \,M$ solution of $NaOH$ for neutralisation. What would be the percentage composition of nitrogen in the compound?
A sample of $0.50 \,g$ of an organic compound was treated according to Kjeldahl?? method. The ammonia evolved was absorbed in $50 \,mL$ of $0.5\, M \,H_{2}SO_{4}$. The residual acid required $60\, mL$ of $0.5 \,M$ solution of $NaOH$ for neutralisation. What would be the percentage composition of nitrogen in the compound?
Updated On: Dec 25, 2024
- $50$
- $60$
- $56$
- $44$
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The Correct Option is C
Solution and Explanation
Given :
Mass of compound taken = $0.50\, g$
Vol. of $H_{2}SO_{4} = 50\, mL$
Molarity of $H_{2}SO_{4} = 0.5\, M$
Vol. of $NaOH$ required = $60 \,mL$
Molarity of $NaOH$ required = $0.5 \,M$
Method adopted : Kjeldahl?? method
Formula Used : $\%$ of $N =\frac{1.4\times M\times2\left[V-\frac{V_{1}}{2}\right]}{m} \ldots\left(i\right)$
By substituting the values in the formula, we get,
$\%$ of $N=\frac{1.4\times0.5\times2\left(50-60 /2\right)}{0.5}=56$
$\therefore\%$ of $N$ in the given compound $=56\%$
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