For the estimation of nitrogen- 1-4 g of organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M10 sulphuric acid- The unreacted acid required 20 mL of M10 sodium hydroxide for complete neutralization- The percentage of nitrogen in the compound is

% of N =1.4×meq. of acidmass of organic compound meq. of H_2SO_4 = 60 ×M10× 2 = 12 meq. of NaOH = 20 ×M10 = 2 ∴ meq. of acid consumed = 12 2 = 10 ∴ % of N = ...

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Standard XII

Chemistry

Organic Compounds

For the estim...

Question

For the estimation of nitrogen, $1.4 g$ of organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in $60 m L$ of $\frac{M}{10}$ sulphuric acid. The unreacted acid required $20 m L$ of $\frac{M}{10}$ sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is

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Solution

$% of N = \frac{1.4 \times meq. of acid}{mass of organic compound}$
$meq. of H_{2} S O_{4} = 60 \times \frac{M}{10} \times 2 = 12 meq. of NaOH = 20 \times \frac{M}{10} = 2$
$∴ meq. of acid consumed = 12-2 = 10$
$∴ % o f N = \frac{1.4 \times 10}{1.4} = 10 %$

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For the estimation of nitrogen, 1.4 g of organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M10 sulphuric acid. The unreacted acid required 20 mL of M10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is

% of N=1.4×meq. of acidmass of organic compound meq. of H2SO4=60×M10×2=12meq. of NaOH=20×M10=2 ∴ meq. of acid consumed = 12-2 = 10 ∴ % of N=1.4×101.4=10%

$% of N = \frac{1.4 \times meq. of acid}{mass of organic compound}$
$meq. of H_{2} S O_{4} = 60 \times \frac{M}{10} \times 2 = 12 meq. of NaOH = 20 \times \frac{M}{10} = 2$
$∴ meq. of acid consumed = 12-2 = 10$
$∴ % o f N = \frac{1.4 \times 10}{1.4} = 10 %$