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Grade: 12th pass
        
An element X binds with oxygen and CO present in air as
 
 
when X at 1 atm is treated with air, 25% of it is bound to CO(g). The partial pressure of CO(g) in air at equilibrium, if partial pressure of O2(g) in air at equilibrium is 0.2 atm, would be
(a) 1.9 × 10–4 atm 
(b) 1.9 × 104 atm 
(c) 2.08 × 10–4 atm 
(d) 2.08 × 104 atm
one month ago

Answers : (1)

Arun
24731 Points
							
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 
−4
 Matm 
−1
 ×0.21atm=2.73×10 
−5
 M
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.
18 days ago
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