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The Henry's law constant for the solubility of Nitrogen gas in water at350 K is8xx 10^(4)atm. The mole fraction of nitrogen in air is0.5.The numberof moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressureis |
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Answer» `4 xx 10 ^(-4)` `(X_(N_(2))) _("in air") = 0.5` TOTAL pressure = 4 atm Partial pressure of nitrogen = Mole fraction `xx` Total pressure `0.5 xx 4 =2` `(P _(N _(2))) = K _(H) xx` Mole fraction of `N_(2)` in solution `2 = 8 xx 10 ^(4) xx ("Number of moles of nitrogen")/("Total number of moles")` `(10 + "No. of moles of" N _(2))/( "No. of moles of" N _(2)) = (8xx 10 ^(4))/(2)` `(10)/("No. of moles of" N_(2)) +1 = 4 xx 10 ^(4)` `(10)/("No. of moles of" N_(2)) = 4000 -1` `THEREFORE ` No. of moles of `N _(2) = (10)/(39999) =2.5 xx 10 ^(-4)` |
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