The Henry's law constant for the solubility of Nitrogen gas iin water at 350 Kis 8xx10^4 atm. The mole fraction of nitrogen in air is 0.5 . The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is …………….... .

The Henry's law constant for the solubility of Nitrogen gas iin water

1.	The Henry's law constant for the solubility of Nitrogen gas iin water at 350 Kis 8xx10^4 atm. The mole fraction of nitrogen in air is 0.5 . The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is …………….... .
Answer» `4xx10^(-4)` `4xx10^(4)` `2xx10^(-2)` `2.5 xx 10^(-4)` Solution :`K_(H) = 8xx10^(4)` `(x_(N_(2)))_("in air")= 0.5` TOTAL pressure = 4 atm Partial pressure of nitrogen = Mole fraction`xx` Total pressure =`0.5xx4 = 2` `(P_(N_(2))) = K_(H)xx "Mole fraction of" N_(2) "in solution"` `2=8xx10^(4)xx("NUMBER of moles of nitrogen")/("Total number of moles")` `(10+"No. of moles of" N_(2))/("No. of moles of" N_(2)) = (8xx10^(4))/(2)` `(10)/("No. of moles of" N_(2))+1 = 4xx10^(4)` `(10)/("No. of moles of" N_(2)) = 40000-1` `therefore` No. of moles of `N_(2) = (10)/(39999) = 2.5xx10^(-4)`