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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 `298K` is `1.0 xx 10^(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 298K and 5 atm pressure is

A

`4.0 xx 10^(-4)`

B

`4.0 xx 10^(-5)`

C

`5.0 xx 10^(-4)`

D

`4.0 xx 10^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(N_(2)) = 0.8 xx 5 = 4 atm`
From Henry law:
`P_(N_(2)) = K X_(N_(2)) [X_(N_(2))` is small]
`4 = 10^(5) xx X_(N_(2))`
`X_(N_(2)) = 4 xx 10^(05)`
`:' X_(N_(2)) lt lt lt n_(H_(2)O) rArr X_(H_(2)O) = 1`
which means if `n_(H_(2)O) =1` than `n_(N_(2)) = 4 xx 10^(-5)`
`rArr` for 10 mole of `H_(2)O` mole of `N_(2)` present are `10 xx 4 xx 10^(-5) = 4 xx 10^(-4)` mole
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