Solve this following


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

(density of water $=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ at $298 \mathrm{~K}$ )

This table implies that:


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

  2. The pressure of a $55.5$ molal solution of $\delta$ is 250 bar

  3. Solubility of $\gamma$ at $308 \mathrm{~K}$ is lower than at $298 \mathrm{~K}$

  4. $\alpha$ has the highest solubility in water at a given pressure

Correct Option: , 2


(1) $\mathrm{P}_{\gamma}=\mathrm{K}_{\mathrm{H}} \mathrm{X}_{\mathrm{Y}}$

$\mathrm{P}_{\gamma}=2 \times 10^{-15} \times \frac{55.5}{55.5+\frac{1000}{18}}=2 \times 10^{-5} \mathrm{~K}$ bar

$=2 \times 10^{-2} \mathrm{bar}$

(2) $\mathrm{P}_{\delta}=\mathrm{K}_{\mathrm{H}} \mathrm{X}_{\delta}$

$\mathrm{P}_{\delta}=0.5 \times \frac{55.5}{55.5+\frac{1000}{18}}=.249 \mathrm{~K}$ bar $=249$ bar

(3) On increasing temperature solubility of gases decreases

(4) $\mathrm{K}_{\mathrm{H}} \downarrow$ solubility $\uparrow$ and lowest $\mathrm{K}_{\mathrm{H}}$ is for $\gamma$.


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