Question

find the mole fraction of O2 in a staurated solutions of oxygen in water at 25°C when the partial pressure of O2 above the water is 0.21 atm. Given that Henry's constant for O2 in water at 25°C is 2.3× 10-51 atm​

Answer 1

Solution:-

☞Total pressure is 1 atm. The mole fraction of oxygen is 0.21.

Hence, the partial pressure of oxygen is 0.21×1=0.21 atm

According to Henry's law, S=KP

S is the solubility in moles per litre

K is the henry's law constant

P is the pressure in atm.

Substitute values in the above expression.

S=1.3×10−4Matm−1×0.21atm=2.73×10−5M

Thus 2.73×10−5 moles of oxygen are dissolved in 1 L of water.

This corresponds to 2.73×10−5×32=8.736×10−4 g or 0.8736 mg of oxygen in 1 liter of solution.

Answer 2

GiVEN

• W = 5g
• WA = 95g
• MB = 60.05gmol^−1
• MA = 18gmol^−1
• P0A = 3.165kPa

Putting the values in expression

PAo − P = MB×WA / WB×MA

3.165 kPa − P = 60.05gmol−1×95g / 5g×18gmol−1

3.165 kPa − P =0.015

P = 3.165kPa − 0.015

P = 3.118kPa