Question

In Kjeldahl’s method for the estimation of nitrogen present in a soil sample, the ammonia evolved from $0.75\text{g}$ of sample neutralized $10\text{mL}$ of $1\text{M}{H}_{2}S{O}_4$. The percentage of nitrogen in the soil is:

• A. $37.33$
• B. $45.33$
• C. $35.33$
• D. $43.33$

Answer 1

The correct option is A $37.33$

This can be solved on the basis of law of chemical equivalance
According to which;
$gramequivalentsofN{H}_3=gramequivalentsof{H}_{2}S{O}_4$
$moles\times nf=\frac{M\times V\times nf}{1000}$
$\to n\times 1=\frac{1\times 10\times 2}{1000}$
$\to n=\frac{20}{1000}=2\times {10}^{-2}$
Now,
1 mole of $N{H}_3$ contains 1 mole of `N`.
$\therefore$ $2\times {10}^{-2}$ moles of $N{H}_3$ contains $2\times {10}^{-2}$ moles of nitrogen.
Hence,
mass of `N `= moles$\times$ molar mass
= $2\times {10}^{-2}\times 14$
= 0.28 g
$\therefore$ % of `N` =$\frac{0.28}{0.75}\times 100$

= $\frac{28}{3}\times 4=$ `37.33 %`