1.

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

Answer»

`4.0xx10^(-4)`
`5.0xx10^(-5)`
`5.0xx10^(-4)`
`4.0xx10^(-6)`

Solution :`P_(N_(2))=K_(H)xx X_(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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