JEE Main 2026 — General Organic Chemistry Question with Solution

Given data:
Mass of organic compound ($w$) = $0.50$ g
Volume of nitrogen gas ($V_1$) = $70$ mL
Temperature ($T_1$) = $300$ K
Total pressure = $715$ mm Hg
Aqueous tension = $15$ mm Hg
Pressure of dry nitrogen ($P_1$) = $715-15=700$ mm Hg
Convert the volume of nitrogen to STP conditions ($P_0=760$ mm Hg, $T_0=273$ K):
Using the combined gas law: $\frac{P_1{V}_1}{T_1}=\frac{P_0{V}_0}{T_0}$
$V_0=\frac{P_1{V}_1{T}_0}{P_0{T}_1}=\frac{700\times 70\times 273}{760\times 300}\approx 58.68$ mL
At STP, $22400$ mL of $N_2$ weighs $28$ g.
Mass of $N_2$ collected = $\frac{28\times 58.68}{22400}$ g
Percentage of nitrogen = $\frac{\text{Mass of }{N}_2}{\text{Mass of compound}}\times 100$
$\%N=\frac{28\times 58.68}{22400\times 0.50}\times 100$
$\%N=\frac{28\times 58.68}{112}=\frac{58.68}{4}=14.67\%$
Rounding to the nearest integer, we get $15$.