Henry's constant (in kbar) for four gases $\alpha, \beta, \gamma$ and $\delta$ in water at 298 K is given below: | | $\alpha$ | $\beta$ | $\gamma$ | $\delta$ | |---|---|---|---|---| | $\mathrm{K}_{\mathrm{H}}$ | 50 | 2 | $2 \times 10^{-5}$ | 0.5 | (density of water $=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ at 298 K ) This table implies that:

(C) The pressure of a 55.5 molal solution of $\delta$ is 250 bar.

(A) $\alpha$ has the highest solubility in water at a given pressure

(B) solubility of $\gamma$ at 308 K is lower than at 298 K .

(D) The pressure of a 55.5 molal solution of $\gamma$ is 1 bar.

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Solutions

Henry's constant (in kbar) for four gases

α, β, γ

and

δ

in water at 298 K is given below:

	$α$	$β$	$γ$	$δ$
$K_{H}$	50	2	$2 \times 1 0^{- 5}$	0.5

(density of water

= 1 0^{3} kg m^{- 3}

at 298 K )

This table implies that:

Answer: (C)

By Henry's law the partial pressure of a dissolved gas is

P = K_{H} X

. A

55.5

molal solution of

δ

has equal moles of gas and of water (

\frac{1000}{18} \approx 55.5

), so its mole fraction is

X = 0.5

; with

K_{H} = 0.5

kbar this gives

P = 0.5 \times 1 0^{3} \times 0.5 = 250

bar, which is the relation the table supports. A lower

K_{H}

would mean higher solubility, so it is

γ

, not

α

, that is the most soluble.

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option