At what partial

pressure oxygen will have a solubility of 0.06 gram per litre in water at 293 Kelvin Henry's law constant of oxygen in water at 303 kelvin is 46.82 k bar  assume the density of the solution to the same as that of water

Asked by pandeyn1604 | 29th May, 2019, 09:16: AM

Expert Answer:

Given:
 
Solubility of O20.06 gram/litre 
 
KH = 46.82 kbar 
 
     = 46.82 × 103 bar
 
Mass of 1 L of water = 1000 gm
 
Mass of solvent = 1000 - 0.06 
 
                       ≈ 1000 gm
 
 
No. of moles of water 
 
 equals space 1000 over 18

equals space 55.55 space mol
 
No. of moles of O2 
 
equals space fraction numerator 0.06 over denominator 16 end fraction

space equals space 0.00375 space mol
 
Mole fraction of O2 
 
space equals space fraction numerator No. of space moles space of space straight O subscript 2 over denominator moles space of space straight O subscript 2 plus space moles space of space straight H subscript 2 straight O end fraction

space equals space fraction numerator 0.00375 over denominator 0.00375 space plus space 55.55 end fraction

space equals space 6.75 cross times 10 to the power of negative 5 end exponent
 
 
We have,
 
PO2 = KH × ΧO2 
 
      = 46.82 × 103 × 6.75 × 10−5 
 
      = 3.160 bar
 
The partial pressure of O2 is 3.16 bar.
 

Answered by Varsha | 29th May, 2019, 12:01: PM