When 1 g of cmnpow1d (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M $$H_{2}SO_{4}$$ was neutrilized by ammonia evolved. The percentage of nitrogen in compound
(X) is :

We need to find the percentage of nitrogen using Kjeldahl's method. Given 1 g of compound, 15 mL of 1 M $$H_2SO_4$$ is neutralised by the evolved $$NH_3$$.

In Kjeldahl's method, nitrogen in the compound is converted to ammonium sulphate, then ammonia is liberated and absorbed in excess $$H_2SO_4$$: $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$

Each mole of $$H_2SO_4$$ reacts with 2 moles of $$NH_3$$ (since $$H_2SO_4$$ is diprotic). Moles of $$H_2SO_4$$ = $$15 \times 10^{-3} \times 1 = 0.015$$ mol, which corresponds to Moles of $$NH_3$$ = $$2 \times 0.015 = 0.03$$ mol. Since each $$NH_3$$ contains one nitrogen atom, moles of N = 0.03 mol.

Substituting into the formula for mass gives $$\text{Mass of N} = 0.03 \times 14 = 0.42 \text{ g}$$

Therefore, the percentage of nitrogen is $$\% N = \frac{0.42}{1} \times 100 = 42\%$$

The correct answer is Option 1: 42.