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

Henry's law constant for `CO_(2)` in water is `2.5xx10^(8)Pa` at `298K`. Calculate mmole of `CO_(2)` dissolved in `14g` water at `2.5atm` pressure at `298K`.
[Take `1atm=10^(5)N//m^(2)` or `Pa`]

A

`12`

B

`8`

C

`4`

D

`2`

Text Solution

Verified by Experts

The correct Answer is:
B
`P_(CO_(2))=k_(H).X_(CO_(2))=K_(H)xx(n_(CO_(2)))/(n_(CO_(2))+N_(water))`
`=K_(H) (n_(CO_(2)))/(n_(water))`
`2.5xx10^(5)=2.5xx10^(8)xx(n_(CO_(2)))/(144//18)`
`n_(CO_(2))=8xx10^(-3)`
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