Calculate the solubility in water in term of mole fraction in partial pressure of `CO_(2)` is `2xx10^(-3)` bar at 298 K temperature, the `K_(a)` value for `CO_(2)` is `6.02xx10^(-4)` bar.

Options

A. `3.322xx10^(-3)`
B. `3.011xx10^(-3)`
C. `3.322xx10^(-4)`
D. `3.011xx10^(-6)`

Answer

C

Explanation

According to Henry.s law `= KH xx CO_(2)`
`X_(CO_(3))=(P_(CO_(3)))/(KH)=(2xx10^(-8)"bar")/(6.02xx10^(-4)"bar")`
`= 3.332xx10^(-4)`.