Henry's constant (in kbar) for four gases $\alpha, \beta, \gamma$ and $\delta$ in water at $298K$ is given below :
$\begin{cases}\alpha & \beta & \gamma & \delta \\ {K}_{H} & 50 & 2 & 2 \times {10}^{- 5} & 0 . 5\end{cases}$
(density of water $= {10}^{3}k{gm}^{- 3}$ at $298K$ ) This table implies that :

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Henry's constant (in kbar) for four gases $\alpha, \beta, \gamma$ and $\delta$ in water at $298K$ is given below :

$\begin{cases}\alpha & \beta & \gamma & \delta \\ {K}_{H} & 50 & 2 & 2 \times {10}^{- 5} & 0 . 5\end{cases}$

(density of water $= {10}^{3}k{gm}^{- 3}$ at $298K$ ) This table implies that :

Quantrex Academy · Accepted Answer

Correct answer: C. The pressure of a 55 . 5 molal solution of is 250 bar

(1) From Henery's law P = k _ H (X) Higher the value of k _ H smaller will be solubility so γ is more soluble. (2) Though solubility of gas will decrease with increase in temperature but this conclusion can not be drawn from the given table. (3)For ⇒ P _ = ( K _ H ) . (X) = 0 . 5 × 10 ^ 3 × [(55 . 55)/(55 . 55 + (1000)/(18))] = 250bar . (4)For γ (p) _ γ = ( K _ H ) _ γ × (x) _ γ = 2 × 10 ^ - 2 [(55 . 5)/(55 . 5 + (1000)/(18))] = 10 ^ - 2 bar

Henry's constant (in kbar) for four gases α, β, γ and in water at...

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