When 1text g$1\text{ g}$ of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15text mL$15\text{ mL}$, 1textM textH_2textSO_4$1\text{M } \text{H}_2\text{SO}_4$ was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is:

### Related Formula \%text N = frac1.4 times M times 2 times VtextMass of compound (g)$$\%\text{ N} = \frac{1.4 \times M \times 2 \times V}{\text{Mass of compound (g)}}$$ Alternatively: textMeq of textH_2textSO_4 = V times M times n_textfactor$$\text{Meq of } \text{H}_2\text{SO}_4 = V \times M \times n_{\text{factor}}$$ ### Core Logic Step 1: Calculate equivalents of acid neutralized: textEquivalents of textH_2textSO_4 = frac15 times 1 times 21000 = 0.03$$\text{Equivalents of } \text{H}_2\text{SO}_4 = \frac{15 \times 1 \times 2}{1000} = 0.03$$ Step 2: Calculate moles of nitrogen atoms: textMoles of textNH_3 = 0.03 implies textMoles of N = 0.03$$\text{Moles of } \text{NH}_3 = 0.03 \implies \text{Moles of N} = 0.03$$ textMass of N = 0.03 times 14 = 0.42text g$$\text{Mass of N} = 0.03 \times 14 = 0.42\text{ g}$$ Step 3: Calculate percentage of nitrogen: \%text N = frac0.421 times 100 = 42\%$$\%\text{ N} = \frac{0.42}{1} \times 100 = 42\%$$ ### Pattern Recognition Sees: Kjeldahl's method with textH_2textSO_4$\text{H}_2\text{SO}_4$ volume and molarity. Shortcut: Use standard formula \%text N = frac1.4 times M times 2 times VW = frac1.4 times 1 times 2 times 151 = 42\%$\%\text{ N} = \frac{1.4 \times M \times 2 \times V}{W} = \frac{1.4 \times 1 \times 2 \times 15}{1} = 42\%$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles & Techniques