Question

^{N_{​​​​​​2} gas is bubbled through water at 298K. How many moles of N_{​​​​​​2} gas would dissolve in 1L water assumed that partial pressure of gas is = 0.987 bar and K_{​​​​​​H} = 76.48 K .}

Answer 1

KH = 76.48kbar = 76480bar

P= 0.989bar

P = KH × x
x= 0.98976480=1.293 ×10−5. This is the mole fraction of N2 gas dissolved in water

Volume of water = 1L = 1000mL
Taking density as 1g/mL, mass of 1L =1000g
Moles of water in 1L = 1000/18=55.56
Let us take the moles of nitrogen as x
Total moles = moles of N2 + moles of water = x + 55.56
mole fraction of N2
1.293 ×10−5= xx+55.561.293 ×10-5x + 1.293 ×10-5 × 55.56 =x x(1-1.293 ×10-5) = 1.293 ×10-5 × 55.56 x=7.17 ×10-4mol = 0.717mmoles