N₂ gas at 293 K, partial pressure = 0.987 bar, KH = 76.48 kbar. Find millimoles dissolved in 1 L water.

Step 1: Apply Henry’s Law (p = KH × x):

x(N₂) = 0.987 / 76480 = 1.290×10⁻⁵

Step 2: Moles of water in 1 L = 1000/18 = 55.5 mol

Step 3: Moles of N₂ (since x << 1):

n(N₂) ≈ x × n(H₂O) = 1.290×10⁻⁵ × 55.5 = 7.16×10⁻⁴ mol = 0.716 mmol

Answer: ≈0.716 millimoles of N₂ dissolve

Key Point: Henry’s Law: gas solubility ∝ partial pressure. The large KH for N₂ means it is only sparingly soluble a key concept in JEE solutions.