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The Henry's law constant fo the solubili...

The Henry's law constant fo 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 298K and 5 atm pressue is

A

`4.0xx10^(-4)`

B

`4.0xx10^(-5)`

C

`5.0xx10^(-4)`

D

`5.0xx10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(N_2)=K_HxxX_(N_2)`
`X_(N_2)=1/(10^5)xx0.8 xx5 = 4xx10^(-5)` per mole In 10 mole, solubility is `4xx10^(-4)`.
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