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

We have an organic sample of mass $$w = 1.4\ \text{g}$$. It is digested according to the Kjeldahl method. The ammonia set free is absorbed in $$60\ \text{mL}$$ of sulphuric acid whose molarity is $$\dfrac{M}{10}=0.1\ M$$.

First we calculate the number of moles of sulphuric acid originally taken. Using the formula

$$\text{Moles} = M \times V(\text{in L})$$

and converting $$60\ \text{mL}$$ to litres, $$V = 60\ \text{mL}=0.060\ \text{L}$$, we get

$$n_{\text{H}_2\text{SO}_4,\ \text{initial}} = 0.1 \times 0.060 = 0.006\ \text{mol}.$$

After absorption of ammonia, some acid remains unreacted. This residual acid is back-titrated with $$20\ \text{mL}$$ of sodium hydroxide of the same molarity $$0.1\ M$$. The moles of sodium hydroxide used are

$$n_{\text{NaOH}} = 0.1 \times \dfrac{20}{1000} = 0.1 \times 0.020 = 0.002\ \text{mol}.$$

For the neutralisation reaction $$\text{H}_2\text{SO}_4 + 2\ \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\ \text{H}_2\text{O}$$, two moles of NaOH correspond to one mole of $$\text{H}_2\text{SO}_4$$. Hence the moles of sulphuric acid left unreacted are

$$n_{\text{H}_2\text{SO}_4,\ \text{left}} = \dfrac{n_{\text{NaOH}}}{2} = \dfrac{0.002}{2} = 0.001\ \text{mol}.$$

The moles of acid that actually reacted with ammonia are therefore

$$n_{\text{H}_2\text{SO}_4,\ \text{reacted}} = n_{\text{H}_2\text{SO}_4,\ \text{initial}} - n_{\text{H}_2\text{SO}_4,\ \text{left}} = 0.006 - 0.001 = 0.005\ \text{mol}.$$

In the Kjeldahl process the neutralisation step is

$$2\ \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4.$$

Thus one mole of $$\text{H}_2\text{SO}_4$$ reacts with two moles of ammonia, and each mole of ammonia contains one mole of nitrogen. So the moles of nitrogen liberated from the compound are

$$n_{\text{N}} = 2 \times n_{\text{H}_2\text{SO}_4,\ \text{reacted}} = 2 \times 0.005 = 0.010\ \text{mol}.$$

The mass of nitrogen present is obtained by multiplying by the molar mass of nitrogen ($$14\ \text{g mol}^{-1}$$):

$$m_{\text{N}} = n_{\text{N}} \times 14 = 0.010 \times 14 = 0.14\ \text{g}.$$

The percentage of nitrogen in the sample is then

$$\%\,\text{N} = \dfrac{m_{\text{N}}}{w} \times 100 = \dfrac{0.14}{1.4} \times 100 = 0.10 \times 100 = 10\%. $$

Hence, the correct answer is Option B.