Question

at what partial pressure of oxygen in solubility of 0.05 gram per litre in water at 293 k

Answer 1

Mass of 1L of solution = 1000mL X 1g mL-1
Mass of solvent = 1000-0.05 ≈1000g
No.of moles of water = 1000/18 = 55.5 mol
No.of moles of nitrogen = 0.05/28 = 1.79 x10-3 mole
mole fraction of N2 = 1.79 x10-3/1.79  x10-3+55.5
=3.22 X 10-5
Partial Presssure Of N2 =kH  X χ N2 =76.48 X103X3.22X10-5
=246.26 bar
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Answer 2

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