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

Options

A. `2.786 g`
B. `0.9273 g`
C. `1.854 g`
D. `3.478 g`

Answer

3

Explanation

Applying Henry's law, we have
`P_(CO_(2))=K_(H)chi_(CO_(2))`
or `chi_(CO_(2))=P_(CO_(2))/K_(H)=(2.5 atm xx101325(Pa)/(atm))/(1.67xx10^(8) Pa)=1.517xx10^(-3)`
By definition
`chi_(CO_(2))=n_(CO_(2))/(n_(CO_(2))+n_(H_(2)O))=1.517xx10^(-3)`
We can find the number of moles of `CO_(2)` in soda water provided we know the number of moles of `H_(2)O`.
Since the solubility of `CO_(2)` in water is very small, we can consider the density of aq. aolution to be `1 g mL^(-1)`. Thus, `500 mL` of `H_(2)O` weighs `500 g`. Now
`n_(H_(2)O)=(mass_(H_(2)O))/(molar mass_(H_(2)O))`
`=(500 g)/(18 g mol^(-1))`
`=27.78 mol`
Substituting this result into mole fraction data, we have
`n_(CO_(2))/(n_(CO_(2))+27.78)~~n_(CO_(2))/27.78=1.517xx10^(-3)`
Thus, `n_(CO_(2))=(27.78 mol)(1.517xx10^(-3))`
`=42.14xx10^(-3) mol`
`=4.214xx10^(-2) mol`
Finally, `mass_(CO_(2))=(n_(CO_(2)))(molar mass_(CO_(2)))`
`=(4.214xx10^(-2) mol)(44 g mol^(-1))`
`=1.854 g`