JEE Chemistry 2026 Question with Solution

Given: Mass of organic compound = $0.50 \, \text{g}$, volume of nitrogen = $70 \, \text{mL}$, temperature = $300 \, \text{K}$, observed pressure = $715 \, \text{mm}$, aqueous tension = $15 \, \text{mm}$.

Find: Percentage of nitrogen in the organic compound.

Step 1: Correct pressure of dry nitrogen gas.

$$P_{\text{dry}} = 715 - 15 = 700 \, \text{mm}$$

Step 2: Convert conditions to STP.

$$V_{\text{STP}} = \frac{70 \times 700 \times 273}{760 \times 300}$$ $$V_{\text{STP}} \approx 56 \, \text{mL}$$

Step 3: Calculate mass of nitrogen.

$$22.4 \, \text{L} \rightarrow 28 \, \text{g}$$ $$56 \, \text{mL} \rightarrow 0.07 \, \text{g}$$

Step 4: Calculate percentage of nitrogen.

$$\%N = \frac{0.07}{0.50} \times 100 = 14$$

The extracted solution working gives $14\%$, but the solution concludes with the correct answer as $18\%$. Therefore, the final recorded answer from the solution is $18\%$.

Given: the solution's shows intermediate working and also explicitly states Correct Answer: 18.

Find: Which answer should be recorded.

The numerical working displayed in the solution leads to:

$$V_{\text{STP}} \approx 56 \, \text{mL}$$

and then to:

$$\%N = 14$$

However, the same the solution explicitly prints Correct Answer: 18 and ends with $18\%$. As per the extraction rule, the solution conclusion is treated for the final answer, while the mismatch is preserved here as a source discrepancy.

Therefore, the final answer is recorded as 18.