Henry's law: mole fractions

  • Chemistry
  • Thread starter Blitzy89
  • Start date
  • #1
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Homework Statement


We know that for dry air, we have:
-78.08% mol N2 ;
-20.94% mol O2.
This is:
- Y(N2) = 0.7808 for nitrogen (the molar fraction in vapor)
- Y (O2) = 0.2094 for oxygen

Knowing that the Henry's law constants are:
for N2 and O2 are 6.51 x 10^7 Torr and 3.3 x 10^7 Torr, respectively

Calculate the mole fractions of N2 and O2 dissolved in water (xN2 and xO2) under total pressure of 1.000 bar


Homework Equations




Henry's law:
The partial pressure = henry's constant k1 x mole fraction of water (x)
P(N2) = K(N2).X(N2)
Same for O2.

Also, from Raoult's law:
Y(N2) = P(N2)/P(tot)
Same for oxygen.

The Attempt at a Solution



I tried applying the equations, but instead of the expected (x(N2)=0.6), I get really small results such as 0.00000651

I am just wondering if I am doing anything wrong by mixing the vapour molar fraction equation with Henry's law for the molar fractions for water?

Thank you.
 

Answers and Replies

  • #2
Borek
Mentor
28,702
3,190
For some reasons I have troubles following what you did and I can't get the picture. But, there are things that are potentially wrong:

Knowing that the Henry's law constants are:
for N2 and O2 are 6.51 x 10^7 Torr and 3.3 x 10^7 Torr, respectively

Henry's law constant is usually listed in pressure/concentration units, these constants have only pressure. What happened to concentration?

And, if the answer is "molar fraction is considered unitless", is it molar fraction of gas in the solution? And if so,

expected (x(N2)=0.6)

what about water, which should have molar fraction well over 0.9?
 
  • #3
11
0
what about water, which should have molar fraction well over 0.9?

Thank you for the tip, this is what I hadn't thought about.
It works out after you include the water.
 

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