the henry's law constant for the solubility of N2 gas in water at 298 K is 1x 105 atm - Chemistry - Solutions

Question

the henry's law constant for the solubility of N2 gas in
water at 298 K is 1x 105 atm the mole of fraction of
N2 in air is 0 8
what is the no of moles of N2 from air dissolved in 10
moles of water of 298 K and 5 atm pressure ??

Geetha · Accepted Answer

pN2 = XN2 × PTot   =  0
8 × 5 = 4 atmFrom Henry's lawpN2 = XN2 × KHXN2 = pN2KH = 41 × 105 = 4 × 10-51 mole solution contain 4 × 10-5 mole of N2 Moles of Water = 1 - 4 × 10-5  = 1 moleThus 1 mole of water contain = 4 × 10-5 mole of N210 moles of water contain= 10 × 4 × 10-5  = 4 × 10-4 mole of N2

the henry's law constant for the solubility of N2 gas in water at 298 K is 1x 105 atm. the mole of fraction of N2 in air is 0.8. what is the no of moles of N2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure ??

the henry's law constant for the solubility of N₂ gas in water at 298 K is 1x 10⁵ atm. the mole of fraction of N₂ in air is 0.8.
what is the no of moles of N₂ from air dissolved in 10 moles of water of 298 K and 5 atm pressure ??