In the Kjedahl’s method for estimation of nitrogen present in soil samples, ammonia evolved from 0.75 g of sample neutralized 10 ml of 1 M ${H_2}S{O_4}$. The percentage of nitrogen in the soil is :
a.) 37.33
b.) 445.33
c.) 35.33
d.) 43.33
Answer
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Hint: The percentage of Nitrogen in Kjeldahl’s method can be given by -
% Nitrogen = $\dfrac{{1.4 \times Normality{\text{ of acid}} \times Volume{\text{ of acid}}}}{{Mass{\text{ of the compound}}}}$
The acid used here is sulphuric acid and the mass of sample 0.75 g is also given. Just by putting the values in formula, we can find the % of Nitrogen in the sample of soil.
Complete answer:
In this type of questions, let us see what is given to us and what we need to find -
Given :
Mass of sample = 0.75 g
Volume of HCl used = 10 ml
Molarity of HCl = 1 M
To solve : Percentage of nitrogen in the soil
We know that the percentage of Nitrogen in Kjeldahl’s method can be given by -
% Nitrogen = $\dfrac{{1.4 \times Normality{\text{ of acid}} \times Volume{\text{ of acid}}}}{{Mass{\text{ of the compound}}}}$
Further, we have 1 M ${H_2}S{O_4}$ = 2 N ${H_2}S{O_4}$
Thus, normality of acid (${H_2}S{O_4}$) = 2
So, putting the values in above formula, we get -
% Nitrogen = $\dfrac{{1.4 \times 2 \times 10}}{{0.75}}$
% Nitrogen = 37.33 %
Thus, the option a.) is the correct answer.
Note:
This method is very useful in determination of nitrogen content. It is used for determination of % of nitrogen in soil samples, waste water sample, fertilizers, meat and feed etc. But it has one limitation. It can not be used for compounds containing nitrogen in azo and nitro groups and even in rings.
% Nitrogen = $\dfrac{{1.4 \times Normality{\text{ of acid}} \times Volume{\text{ of acid}}}}{{Mass{\text{ of the compound}}}}$
The acid used here is sulphuric acid and the mass of sample 0.75 g is also given. Just by putting the values in formula, we can find the % of Nitrogen in the sample of soil.
Complete answer:
In this type of questions, let us see what is given to us and what we need to find -
Given :
Mass of sample = 0.75 g
Volume of HCl used = 10 ml
Molarity of HCl = 1 M
To solve : Percentage of nitrogen in the soil
We know that the percentage of Nitrogen in Kjeldahl’s method can be given by -
% Nitrogen = $\dfrac{{1.4 \times Normality{\text{ of acid}} \times Volume{\text{ of acid}}}}{{Mass{\text{ of the compound}}}}$
Further, we have 1 M ${H_2}S{O_4}$ = 2 N ${H_2}S{O_4}$
Thus, normality of acid (${H_2}S{O_4}$) = 2
So, putting the values in above formula, we get -
% Nitrogen = $\dfrac{{1.4 \times 2 \times 10}}{{0.75}}$
% Nitrogen = 37.33 %
Thus, the option a.) is the correct answer.
Note:
This method is very useful in determination of nitrogen content. It is used for determination of % of nitrogen in soil samples, waste water sample, fertilizers, meat and feed etc. But it has one limitation. It can not be used for compounds containing nitrogen in azo and nitro groups and even in rings.
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