00:01
Let's calculate the solubility 20 degrees celsius, which is 293 kelvin.
00:10
So, lon, x2, which would be a mole fraction, is equal to negative delta h formation over r, 1 over t, minus 1 over tf.
00:23
This is lon x2, 0 .13 .22, 10 to the 3 joule per mole.
00:30
R's our ideal gas constant, a .314, joule per mole kelvin, 1 over 293 kelvin minus 1 over 359 .9 kelvin.
00:46
Solving for x2 here is 0 .365.
00:52
Now to get our solubility here.
00:54
This would be 0 .365, total 1 minus 0 .365...