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The Henry's law constant for the solubil...

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`.

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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

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 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 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 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 constant for solubility of N_(2) gas in water at 298K 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 pressure is :

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 mole of water at 298 K and 5 atm pressure is: