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The henry's law constant for the solubil...

The henry's law constant for the solubility of `N_2` gas in water at 298 K is `1.0xx10^5` atm . The mole fraction of `N_2 ` in air is 0.8 . The number of moles of `N_2` from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

A

`4.0xx10^(-4)`

B

`4.0xx10^(-5)`

C

`5.0xx10^(-4)`

D

`4.0xx10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
(A)

`P=K_(H)chi_(N_(2))`
`0.8xx5=1xx10^(5)xx chi_(N_(1))`
`chi_(N_(2))=4xx10^(-5)` (in 10 moles of water)
`rArr 4xx10^(-5)=(n_(N_(1)))/(n_(N_(1))+10)`
`n_(N_(1))xx5xx4xx10^(-4)=n_(N_(2))`
`rArrn_(N_(1))=4xx10^(-4)`
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