TG EAMCET 3 May 2025 Shift 2 Paper

The organic compound $X$ is $\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2$ (aniline). Here's a concise breakdown of the solution.We're using the Kjeldahl method to find the nitrogen content of compound $X$.(A) Moles of acid reacted with ammoniaTotal $\mathrm{H}_2\mathrm{SO}_4$ added ; $0.060\ \mathrm{L} \times 0.1\ \mathrm{M} = 0.006\ \mathrm{mol}$.NaOH used in back-titration : $0.020\ I \times 0.1\ \mathrm{M} = 0.002\ \mathrm{mol}$.Unreacted $\mathrm{H}_2\mathrm{SO}_4$ (from $2\ \mathrm{NaOH} : 1\ \mathrm{H}_2\mathrm{SO}_4$ ) : $\frac{0.002\ \mathrm{mol}}{2} = 0.001\ \mathrm{mol}$.$\mathrm{H}_2\mathrm{SO}_4$ reacted with $\mathrm{NH}_3: 0.006\ \mathrm{mol}\ 0.001\ \mathrm{mol} = 0.005\ \mathrm{mol}$.(B) Mass of nitrogen in compound $X$Moles of $\mathrm{NH}_3$ (from $2\ \mathrm{NH}_3$ :$1\ \mathrm{H}_2\mathrm{SO}_4$ ) : $0.005\ \mathrm{mol} \times 2 = 0.010\ \mathrm{mol}$.Mass of nitrogen ( $N = 14.01\ \mathrm{g/mol}$ ) : $0.010\ \mathrm{mol} \times 14.01\ \mathrm{g/mol} = 0.1401\ \mathrm{g}$.(C) Percentage of nitrogen in compound $X$.Sample mass $= 0.933\ \mathrm{g}$$\% N = \left( \frac{0.1401\ \mathrm{g}}{0.933\ \mathrm{g}} \right) \times 100\%$$\approx 15.01\%$.(D) Compare with optionsAniline $(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2)$ : Contains one N .Molar mass $\approx 93\ \mathrm{g/mol}. \% N = \left( \frac{14}{93} \right) \times 100\% \approx 15.05\%$.Other options (benzylamine, n-propylamine, acetamide) have significantly different nitrogen percentages (approx. $13.08\%, 23.73\%, 23.73\%$ respectively).Since compound $X$ has approximately $15.01\%$ nitrogen, it matches Aniline.