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:

`4.0xx10^(-4)`

`4.0xx10^(-5)`

`5.0xx10^(-4)`

`5.0xx10^(-5)`

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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

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

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 0.6 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm pressure is a. 3.0 xx 10^(-4) b. 4.0 xx 10^(-5) c. 5.0 xx 10^(-4) d. 6.0 xx 10^(-6)

The Henry's law constant for the solubility of N_(2) gas in water at 298 K is 1xx10^(-5)" atm" . The mole fraction of N_(2) in air in 0.8. If the number of moles of N_(2) of air dissolved in 10 moles of water at 298 K and 5 atm x. 10^(-4) . Find the value of x.

The Henry's law constant for the solubility of N_(2) gas i water at 298K is 1xx10^(5) atm. The mole fraction of N_(2) air is 0.8 . The number of mole of N_(2) of dissolved in 10 mole of water at 298K and 5 atm. are x xx10^(4) . Find the value of x .

Henry's law constant for the solubility of nitrogen gas in water at 298 K is 1.0 xx 10^(-5) atm . The mole fraction of nitrogen in air is 0.8 .The number of moles of nitrogen from air dissolved in 10 mol of water at 298 K and 5 atm pressure is

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