The quantity of CO_2 in 500 mL of soda water | vedclass

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(A) Given: Pressure of $CO_2$ $(p)$ $= 3.34 \ bar = 3.34 	imes 10^5 \ Pa$. Henry's law constant $(K_H)$ for $CO_2$ at $298 \ K$ is $1.67 	imes 10^8

Vedclass · Accepted Answer

$2.442$

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(A) Given: Pressure of $CO_2$ $(p)$ $= 3.34 \ bar = 3.34 	imes 10^5 \ Pa$. Henry's law constant $(K_H)$ for $CO_2$ at $298 \ K$ is $1.67 	imes 10^8 \ Pa$. According to Henry's law,$p = K_H 	imes x$,where $x$ is the mole fraction of $CO_2$. $x = \frac{p}{K_H} = \frac{3.34 	imes 10^5 \ Pa}{1.67 	imes 10^8 \ Pa} = 2 	imes 10^{-3}$. Volume of water $= 500 \ mL$. Since density of water is $1 \ g/mL$,mass of water $= 500 \ g$. Moles of water $(n_{H_2O})$ $= \frac{500 \ g}{18 \ g/mol} = 27.78 \ mol$. Since $x = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}}$,we have $n_{CO_2} = x 	imes n_{H_2O} = 2 	imes 10^{-3} 	imes 27.78 \ mol = 0.05556 \ mol$. Mass of $CO_2 = n_{CO_2} 	imes 	ext{Molar mass of } CO_2 = 0.05556 \ mol 	imes 44 \ g/mol = 2.4446 \ g \approx 2.442 \ g$.

The quantity of $CO_2$ in $500 \ mL$ of soda water when packed under $3.34 \ bar$ $CO_2$ pressure at $298 \ K$ in $g$ is

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