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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

`4.0xx10^(-4)`

B

`4.0xx10^(-5)`

C

`5.0xx10^(-4)`

D

`4.0xx10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
(A)
`P=K_(H)chi_(N_(2))`
`0.8xx5=1xx10^(5)xx chi_(N_(1))`
`chi_(N_(2))=4xx10^(-5)` (in 10 moles of water)
`rArr 4xx10^(-5)=(n_(N_(1)))/(n_(N_(1))+10)`
`n_(N_(1))xx5xx4xx10^(-4)=n_(N_(2))`
`rArrn_(N_(1))=4xx10^(-4)`
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Knowledge Check

  • 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 dissolve in 10 moles of water at 298 K and 5 atm pressure is

    A
    `4 xx 10^(-4)`
    B
    `4.0 xx10^(-5)`
    C
    `5 xx 10^(-4)`
    D
    `4.0 xx 10^(-6)`

  • 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)`

  • 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
    `4.0xx10^(-4)`
    B
    `5.0xx10^(-5)`
    C
    `5.0xx10^(-4)`
    D
    `4.0xx10^(-6)`

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