Henry's constant (in kbar) for four gases alp | vedclass

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(B) According to Henry's Law, $P = K_{H} \cdot X$, where $X$ is the mole fraction of the gas.For a $55.5 \ m$ solution, there are $55.5 \ mol$ of solu

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) According to Henry's Law, $P = K_{H} \cdot X$, where $X$ is the mole fraction of the gas.For a $55.5 \ m$ solution, there are $55.5 \ mol$ of solute in $1000 \ g$ of water.Moles of water $= \frac{1000 \ g}{18 \ g/mol} \approx 55.5 \ mol$.Mole fraction $X = \frac{55.5}{55.5 + 55.5} = 0.5$.For $\delta$: $P_{\delta} = K_{H} \cdot X = 0.5 \ kbar 	imes 0.5 = 0.25 \ kbar = 250 \ bar$.For $\gamma$: $P_{\gamma} = 2 	imes 10^{-5} \ kbar 	imes 0.5 = 1 	imes 10^{-5} \ kbar = 0.01 \ bar$.Since $K_{H}$ is inversely proportional to solubility, $\gamma$ has the highest solubility, not $\alpha$.Solubility of gases generally decreases with an increase in temperature, so the statement in option $C$ is incorrect.Thus, option $B$ is the only correct statement.

Gas	$K_{H} \ (kbar)$
$\alpha$	$50$
$\beta$	$2$
$\gamma$	$2 \times 10^{-5}$
$\delta$	$0.5$

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