Q25 / 180 · KCET 2010 · 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

  • Correct answer
    A
    $0.84$
  • B
    $8.4$
  • C
    $16.8$
  • D
    $1.68$

Solution

From Kjeldahl's method, percentage of nitrogen $$ \begin{aligned} &=\frac{1.4 \times \mathrm{N} \times \mathrm{V}}{\mathrm{w}}=\frac{1.4 \times 0.1 \times 30}{5} \\ &=0.84 \% \end{aligned} $$

Hence, the correct answer is option: 'A'.

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