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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
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Verified Answer
The correct answer is:
$0.84$
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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