Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from $0.55 g$ of the compound neutralised $12.5 m L$ of $1 M H_{2} S O_{4}$ solution. The percentage of nitrogen in the compound is (Nearest integer)

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Solution

Meq. of $H_{2} S O_{4}$ = Meq. of $N H_{3}$
12. 5 × 1 × 2 = 25 meq. of $N H_{3}$
= 25 millimoles of $N H_{3}$ .
So Millimoles of 'N' = 25
Moles of 'N' = $25 \times 10^{- 3}$
wt. of N = $14 \times 25 \times 10^{- 3}$

$% N = \frac{1.4 \times N \times V}{Mass of organic compound}$

$= \frac{1.4 \times 2 \times 12.5}{0.55} = 63.63 % \approx 64$

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Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55 g of the compound neutralised 12.5 mL of 1 M H2SO4 solution. The percentage of nitrogen in the compound is (Nearest integer)

Meq. of H2SO4 = Meq. of NH3 12. 5 × 1 × 2 = 25 meq. of NH3 = 25 millimoles of NH3. So Millimoles of 'N' = 25 Moles of 'N' = 25×10–3 wt. of N = 14×25×10–3 % N=1.4×N×VMass of organic compound =1.4×2×12.50.55=63.63% ≈64

Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from $0.55 g$ of the compound neutralised $12.5 m L$ of $1 M H_{2} S O_{4}$ solution. The percentage of nitrogen in the compound is (Nearest integer)