At 298 K,how many millimoles of CO_2 gas wil | vedclass

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(D) According to Henry's Law,the mole fraction of $CO_2$ is given by: $x_{CO_2} = \frac{P_{CO_2}}{K_H} = \frac{2 	imes 10^{-8}}{6.02 	imes 10^{-4}}

Vedclass · Accepted Answer

$1.661$

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(D) According to Henry's Law,the mole fraction of $CO_2$ is given by: $x_{CO_2} = \frac{P_{CO_2}}{K_H} = \frac{2 	imes 10^{-8}}{6.02 	imes 10^{-4}} = 3.322 	imes 10^{-5}$ Since the density of water is $1 \ g \ mL^{-1}$,the mass of $900 \ mL$ of water is $900 \ g$. The number of moles of water $(n_{H_2O})$ is: $n_{H_2O} = \frac{900 \ g}{18 \ g \ mol^{-1}} = 50 \ mol$ Since $x_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}}$,and $n_{CO_2}$ is very small compared to $n_{H_2O}$,we can approximate $n_{CO_2} + n_{H_2O} \approx n_{H_2O} = 50 \ mol$. $3.322 	imes 10^{-5} = \frac{n_{CO_2}}{50}$ $n_{CO_2} = 3.322 	imes 10^{-5} 	imes 50 = 1.661 	imes 10^{-3} \ mol$ Converting to millimoles: $1.661 	imes 10^{-3} \ mol 	imes 10^3 \ mmol \ mol^{-1} = 1.661 \ mmol$ Thus,$1.661 \ mmol$ of $CO_2$ gas will dissolve.

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

At $298 \ K$,how many millimoles of $CO_2$ gas will dissolve in $900 \ mL$ of water when $CO_2$ gas is passed through it? The $K_H$ value is $6.02 \times 10^{-4} \ bar$ and the partial pressure of $CO_2$ gas is $2 \times 10^{-8} \ bar$.

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