JEE Main 2025 — 2 April, Evening Shift — Question 10

Dry pressure of nitrogen, $P = 715\ \text{mm Hg} - 15\ \text{mm Hg} = 700\ \text{mm Hg}$

Volume of nitrogen, $V = 60\ \text{mL} = 0.060\ \text{L}$

Temperature, $T = 300\ \text{K}$

Gas constant, $R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$

Convert pressure to atm, we have

$$700\ \text{mm Hg} = \frac{700}{760}\ \text{atm}$$

Moles of nitrogen:

$$n = \frac{PV}{RT} = \frac{\left(\frac{700}{760}\right)\times 0.060}{0.0821 \times 300}$$ $$n \approx 0.00224\ \text{mol}$$

Mass of nitrogen:

$$m = n \times 28 = 0.00224 \times 28 = 0.0627\ \text{g}$$

Percentage of nitrogen:

$$\%\ \text{N} = \frac{0.0627}{0.5} \times 100 = 12.54 \approx 12.57$$

Thus, the correct answer is A.