00:01
We have the given values.
00:02
So basically we have our kh that is equals to 0 .00130 molar concentration per atmospheric.
00:12
We have our p, that is 21%, 1 atm.
00:19
We have also 0 .21 times 1 atm.
00:24
We have p is equals to 0 .21 atm.
00:28
And c is equals to kh.
00:31
P.
00:33
So basically here we have equal to 0 .00130 molar per atmospheric pressure times 0 .218m.
00:45
So we have c is equals to 0 .000273m.
00:50
Now the solubility of oxygen in water when exposed to air at 1 .0080m is 0 .00273.
00:58
So the solubility of oxygen is obtained by substituting the given hand risk law constant and pressure in hand risk formula.
01:06
So the result on solubility is obtained in mole per liter.
01:10
Now let's go to question two.
01:12
Now for question two, we have also the given values for our kh is 0 .00130.
01:19
Our p is 21 % that is 0 .89.
01:23
So to get our p we will just multiply 0 .21 times 0 .89 and this would be p is equals to 0 .1869.
01:33
Atm.
01:35
So we were going to use the formula that it c is equals to khp...