Question

6. If the solubility of O2 in water at 25°C when PO2 = 0.360 atm is 15.0 g/100 g H2O, what is the solubility of O2 at 25°C when PO2 is 1.72 atm? A) 71.7 g/100 g H2O B) 24.2 g/100 g H2O C) 0.319 g/100 g H2O D) 0.0140 g/100 g H2O E) 3.14 g/100 g H2O

Answer 1

First determine the value of Henry's constant for O2(g):

Pressure is given in atm, convert it into kpa,

0.360 atm = 36.477 kpa

1.72 atm = 174.27

S = kP
k = S/P = (0.015 g/L)/(36.477 kPa) = 0.0004112 g/LkPa

now calculate the solubility of O2(g) at the new pressure:

S = kP = (0.0004112 g/LkPa)(174.27 kPa) = 0.07165 g/litre

This value is in gm/litre,

Now take it for 100 gm(same as ml)

71.65 gm/100gm. Which is nothing but 71.5 g/100g.

Hope this will help you. If you are satisfied with my answer, please leave your positive feedback.

Thanks.