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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

`4xx10^(-4)`

B

`4.0xx10^(-5)`

C

`5.0xx10^(-4)`

D

`4.0xx10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(N_(2))=K_(H)xxx_(N_(2))`
`x_(N_(2))=(1)/(10^(5)xx0.8xx5=4xx10^(-5)` per mole
In `10` mole solubility is `4xx10^(-4)`.
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