Question

# The Henry's law constant for O2 in water at 25 °C is 1.25 × 10-5 mol...

The Henry's law constant for O2 in water at 25 °C is 1.25 × 10-5 mol L-1 kPa-1, and the solubility is 1.5 × 10-3 mol L-1. What is the partial pressure of O2 under these conditions?

#### Homework Answers

Answer #1

Given : Henry's law constant for O2 in water at 25 °C is (k)1.25 × 10-5 mol L-1 kPa-1

solubility is (Cg) = 1.5 × 10-3 mol L-1

According to the Henry's law

For many gaseous solutes, the relation between solubility, Cg, and partial pressure, Pg, is a proportional one: Cg = k* Pg

where k is a proportionality constant that depends on the identities of the gaseous solute and solvent

Cg = k* Pg

1.5 × 10-3 mol L-1 = 1.25 × 10-5 mol L-1 kPa-1 *  Pg

Pg = (1 .5 × 10-3 mol L-1 ) / (1.25 × 10-5 mol L-1 kPa-1) = 120 kPa

The partial pressure of O2 under these conditions = 120 kPa

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