1.

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.

Answer»

`3.322xx10^(-3)`
`3.011xx10^(-3)`
`3.322xx10^(-4)`
`3.011xx10^(-6)`

Solution :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)`.


Discussion

No Comment Found

Related InterviewSolutions