Henry's constant (in kbar) for four gases , , and in water at 298 K is given below: | | | | | | |---|---|---|---|---| |…

Strategy:\n> Henry's law constant ($K_H$) inversely relates to solubility. For a given pressure, the gas with the lowest $K_H$ will have the highest mole fraction (solubility). We also check the pressure for a 55.5 molal solution (which means $x = 0.5$).\n\nStep 1: Evaluate Statement (3)\nThe statement mentions a 55.5 molal solution of $\delta$.\n- 55.5 mol of solute per 1 kg (55.5 mol) of water.\n- Mole fraction $x = \\frac{55.5}{55.5 + 55.5} = 0.5$.\nUsing Henry's law $p = K_H \cdot x$ for $\delta$ ($K_H = 0.5\\text{ kbar}$):\n$p = 0.5\\text{ kbar} \\times 0.5 = 0.25\\text{ kbar} = 250\\text{ bar}$\nThis matches statement (3) exactly.\n\nStep 2: Determine why other statements are incorrect\n- (a): $\alpha$ has the highest $K_H$ (50 kbar), which means it has the lowest solubility.\n- (d): For $\gamma$, $K_H = 2 \\times 10^{-5}\\text{ kbar}$. At $x=0.5$:\n $p = (2 \\times 10^{-5}) \\times 0.5 = 10^{-5}\\text{ kbar} = 0.001\\text{ bar} \neq 1\\text{ bar}$.\n\n$\boxed{\\text{Answer: (3)}}$