Question

# The solubility of oxygen, O2, in water is 7.20 ✕ 10−4 mol/L at 0°C when the...

The solubility of oxygen, O2, in water is 7.20 ✕ 10−4 mol/L at 0°C when the nitrogen pressure above water is 0.554 atm. Calculate the solubility of oxygen in water when the partial pressure of oxygen above water is 1.083 atm at 0°C? The Henry's Law constant for oxygen is 1.30 ✕ 10−3 mol/ L atm.

#### Homework Answers

Answer #1

Henry's Law expression:

Solubility of a gas (S) = K Pgas ....where K = Henry's law constant, P=pressure of a gas above solution, S=solubility in mol/L

Given, K = 1.30*10^-3 mol/Latm, PO2 = 1.083 atm, S= ?

S = (1.3*10^-3)*1.083 = 1.41*10^-3 mol/L

Thus the solubility of O2 at 1.083 atm and 0oC = 1.41*10^-3 mol/L

Please note that the other information may be given to calculate Henry's law constant. But I am not sure about nitrogen, I think it is typographic, instead of oxygen they gave nitrogen. Because if we calculate K using  7.20 ✕ 10−4 mol/L and 0.554 atm,

K = 1.3*10^-3 mol/Latm.

Know the answer?
Your Answer:

#### Post as a guest

Your Name:

What's your source?

#### Earn Coins

Coins can be redeemed for fabulous gifts.

##### Not the answer you're looking for?
Ask your own homework help question
ADVERTISEMENT
##### Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

ADVERTISEMENT