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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 `1.0 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 `298 K` and `5 atm`. Pressure is:

A

`4.0 xx 10^(-4)`

B

`4.0 xx 10^(a)`

C

`5.0 xx 10^(-4)`

D

`4.0 xx 10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`:' P_(N_(2)) = K_(H) xx X_(N_(2))`
`:. X_(N_(2)) = (0.8 xx 5)/(1.0 xx 10^(5)) = 4.0 xx 10^(-5) mol^(-1)`
`because (n_(N_(2)))/(n_(N_(2)) + n_(H_(2))O) = X_(N_(2)) :. (n_(N_(2)))/(n_(N_(2)) + 10) = 4.0 xx 10^(-5)`
`rArr n_(N_(2)) ~~ 4.0 xx 10^(-4)`
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