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Henry's law constant for CO(2) in water ...

Henry's law constant for `CO_(2)` in water is `1.67xx10^(8)` Pa at 298 K. Calculate the quantity of `CO_(2)` in 500 mL of soda water when packed under 2.5 atm `CO_(2)` pressure at 298 K.

Text Solution

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Step I : Calculation of number of moles of `CO_(2)`
According to Henry's law,
Mole fraction of `CO_(2)(x_(CO_(2)))=("Partial pressure of "CO_(2))/(K_(H)" for "CO_(2))`
`K_(H)=1.67xx10^(8)" Pa"=((1" atm"))/((101325"Pa"))xx(1.67xx10^(8)"Pa")`
`=1.648xx10^(3)" atm"`
`x_(CO_(2))=(2.5" atm")/((1.648xx10^(3)" Pa"))=1.52xx10^(-3)`
`n_(H_(2)O)=((500"g"))/((18" g mol"^(-1)))=27.78" mol"`
`x_(CO_(2))=(n_(CO_(2)))/(n_(CO_(2))+n_(H_(2)O))=(n_(CO_(2)))/(n_(H_(2)O))=(n_(CO_(2)))/((27.78" mol"))`
`n_(CO_(2))=x_(CO_(2))xx(27.78" mol")=(1.52" "10^(-3))xx(27.78" mol")=0.0422" mol"`
Note : `n_(CO_(2))` is considered negligible due to its little solubility in water.
Step II : Calculation of mass of dissolved `CO_(2)` in water
Mass of `CO_(2)=" No. of moles "CO_(2)xx" Molar mass of "CO_(2)`
`=(0.0422" mol")xx(44" g mol"^(-1))=1.857" g"`
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