KCET · Chemistry · General Organic Chemistry
In Kjeldahl's method, ammonia from \(5 \mathrm{~g}\) of food neutralizes \(30 \mathrm{~cm}^{3}\) of \(0.1 \mathrm{~N}\) acid. The percentage of nitrogen in the food is
- A \(0.84\)
- B \(8.4\)
- C \(16.8\)
- D \(1.68\)
Answer & Solution
Correct Answer
(A) \(0.84\)
Step-by-step Solution
Detailed explanation
From Kjeldahl's method, percentage of nitrogen
\[
\begin{aligned}
&=\frac{1.4 \times \mathrm{N} \times \mathrm{V}}{\mathrm{w}}=\frac{1.4 \times 0.1 \times 30}{5} \\
&=0.84 \%
\end{aligned}
\]
\[
\begin{aligned}
&=\frac{1.4 \times \mathrm{N} \times \mathrm{V}}{\mathrm{w}}=\frac{1.4 \times 0.1 \times 30}{5} \\
&=0.84 \%
\end{aligned}
\]
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