If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 - Chemistry - Solutions

Question

If N2 gas is bubbled through water at 293 K, how many millimoles
of N2 gas would dissolve in 1 litre of water Assume that N2 exerts
a partial pressure of 0 987 bar Given that Henrys law constant for
N2 at 293 K is 76 48 kbar

veena Vasandani · Accepted Answer

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 56 1
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

If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henrys law constant for N2 at 293 K is 76.48 kbar.