If the solubility of H_2S in water at STP is | vedclass

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(D) The molality $m$ is given by $m = \frac{n_{solute}}{W_{solvent} 	ext{ (in kg)}}$.Given $m = 0.195 \, mol/kg$,this means $0.195 \, mol$ of $H_2S$

Vedclass · Accepted Answer

$285$

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(D) The molality $m$ is given by $m = \frac{n_{solute}}{W_{solvent} 	ext{ (in kg)}}$.Given $m = 0.195 \, mol/kg$,this means $0.195 \, mol$ of $H_2S$ is dissolved in $1000 \, g$ $(1 \, kg)$ of water.Number of moles of water $n_{H_2O} = \frac{1000 \, g}{18 \, g/mol} = 55.55 \, mol$.Mole fraction of $H_2S$ $(X_{H_2S})$ = $\frac{n_{H_2S}}{n_{H_2S} + n_{H_2O}} = \frac{0.195}{0.195 + 55.55} = \frac{0.195}{55.745} \approx 0.0035$.According to Henry's Law,$P = K_H 	imes X$.At $STP$,pressure $P = 0.987 \, bar \approx 1 \, bar$.$K_H = \frac{P}{X_{H_2S}} = \frac{1}{0.0035} \approx 285.7 \, bar$.Thus,the value is approximately $285 \, bar$.

Gas	$K_H$ (kbar at $298 \ K$)
$CO_2$	$1.67$
$Ar$	$40.3$
$HCHO$	$1.83 \times 10^{-5}$
$CH_4$	$0.413$

If the solubility of $H_2S$ in water at $STP$ is $0.195 \, m$,what is the value of $K_H$ in $bar$?

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