Question:
In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from $0.75\, g$ of sample neutralized $10\, mL$ of ${1 \, M \, H_2SO_4}$. The percentage of nitrogen in the soil is :
In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from $0.75\, g$ of sample neutralized $10\, mL$ of ${1 \, M \, H_2SO_4}$. The percentage of nitrogen in the soil is :
Updated On: May 9, 2026
- 37.33
- 45.33
- 35.33
- 43.33
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The Correct Option is A
Solution and Explanation
$\because \; M \times V (ml) = m \; mol$
$10 \; m \; mol \; {H_2SO_4 } = 20 \; m \; mol \; of \; {NH_3}$
${[H_2SO_4 + 2NH_3 -> (NH_4)_2 SO_4]}$
${1 \; mol \; NH_3}$ contains 14 g nitrogen
${20 \times 10^{-3} \; mol \; NH_3}$ contains ${14 \times 20 \times 10^{-3}}$ nitrogen
0.75 g of sample contains
$\% $ Nitrogen $ = \frac{14 \times 20 \times 10^{-3}}{0.75} \times 100 = 37.33 \%$
$10 \; m \; mol \; {H_2SO_4 } = 20 \; m \; mol \; of \; {NH_3}$
${[H_2SO_4 + 2NH_3 -> (NH_4)_2 SO_4]}$
${1 \; mol \; NH_3}$ contains 14 g nitrogen
${20 \times 10^{-3} \; mol \; NH_3}$ contains ${14 \times 20 \times 10^{-3}}$ nitrogen
0.75 g of sample contains
$\% $ Nitrogen $ = \frac{14 \times 20 \times 10^{-3}}{0.75} \times 100 = 37.33 \%$
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