0.8 g of an organic compound was analysed by Kjeldahl's method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42%, then _________ mL of 1M H$$_2$$SO$$_4$$ would have been neutralized by the ammonia evolved during the analysis.

In Kjeldahl’s method, the nitrogen present in the organic compound is finally converted into ammonia. That ammonia is absorbed by a known, measured volume of standard acid. By knowing how much of the acid is neutralised we can back-calculate the amount of nitrogen. Here we will move in the reverse direction: we already know the mass of nitrogen, and we have to find the volume of 1 M $$\mathrm{H_2SO_4}$$ that would be neutralised.

We are given that $$0.8\ \text{g}$$ of the organic compound contains $$42\%$$ nitrogen by mass. First we calculate the actual mass of nitrogen present:

$$\text{Mass of nitrogen}=0.8\ \text{g}\times \dfrac{42}{100}=0.336\ \text{g}$$

Next we convert this mass of nitrogen to moles. The atomic mass of nitrogen is $$14\ \text{g mol}^{-1}$$, so

$$\text{Moles of }N=\dfrac{0.336\ \text{g}}{14\ \text{g mol}^{-1}}=0.024\ \text{mol}$$

In the Kjeldahl procedure each atom of nitrogen becomes one molecule of ammonia, $$\mathrm{NH_3}$$. Therefore,

$$\text{Moles of }NH_3=\text{Moles of }N=0.024\ \text{mol}$$

Now we write the neutralisation reaction between ammonia and sulphuric acid:

$$2\,\mathrm{NH_3}+ \mathrm{H_2SO_4}\;\longrightarrow\; (\mathrm{NH_4})_2\mathrm{SO_4}$$

From the balanced equation, $$2$$ moles of $$\mathrm{NH_3}$$ require $$1$$ mole of $$\mathrm{H_2SO_4}$$. Hence the moles of $$\mathrm{H_2SO_4}$$ needed are

$$\text{Moles of }H_2SO_4=\dfrac{\text{Moles of }NH_3}{2}=\dfrac{0.024}{2}=0.012\ \text{mol}$$

The concentration (molarity) of the acid is given as $$1\ \text{M}$$, meaning

$$\text{Molarity}=\dfrac{\text{Moles of solute}}{\text{Volume of solution in litres}}$$

Re-arranging for volume, we get

$$\text{Volume}=\dfrac{\text{Moles}}{\text{Molarity}}=\dfrac{0.012\ \text{mol}}{1\ \text{mol L}^{-1}}=0.012\ \text{L}$$

Converting litres to millilitres ($$1\ \text{L}=1000\ \text{mL}$$):

$$0.012\ \text{L}\times1000=12\ \text{mL}$$

So, the answer is $$12$$.