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,the number of moles of solute $=

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

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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,the number of moles of solute $= 55.5 \ mol$ 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 $\delta$: $P_{\delta} = K_{H} \cdot X = 0.5 \ kbar 	imes 0.5 = 0.25 \ kbar = 250 \ bar$.Thus,option $B$ is correct.For $\gamma$: $P_{\gamma} = 2 	imes 10^{-5} \ kbar 	imes 0.5 = 1 	imes 10^{-5} \ kbar = 0.01 \ bar$.Solubility of gases generally decreases with an increase in temperature,so option $C$ is correct.Since $K_{H}$ is inversely proportional to solubility,$\gamma$ has the highest solubility,not $\alpha$.

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

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

Gas $K_{H} \ (kbar)$

$\alpha$ $50$

$\beta$ $2$

$\gamma$ $2 \times 10^{-5}$

$\delta$ $0.5$

(Density of water $= 10^{3} \ kg \ m^{-3}$ at $298 \ K$). This table implies that:

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