11. If O2 gas is bubbled through water at 293 K, how many centimoles of O2 gas would dissolve in 1 L of water? Assume that O2 gas exerts a partial pressure of 0.823 bar. Given that the Henry’s law constant for O2 at 293 K is 34.86 kbar.
Answers
Answer:
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Explanation:
According to Henry's law, P=Kx
x is the mole fraction
K is the henry's law constant
P is the pressure in atm.
Substitute values in the above expression.
x= 76480 bar
0.987 bar
=1.29×10 −5
= 55.55 n
n=7.16×10 −4 = 1mol
7.16×10 −4
mol×1000 mmol
=0.716 mmol
Answer:
Explanation:
The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law.
x(oxygen) = p(oxygen)/KH=0.987 bar/34860 bar= 2.8313 × 10−5
As 1 litre of water contains 55.5mol of it, therefore if n represents number of moles of O2 in solution
x(oxygen) = p(oxygen)KH=0.987 bar34860 bar= 2.8313 × 10-5
n is very small, so n+55.5 ≈ 55.5
So,n=2.83× 10−5 ×55.5 =0.001571 =15.71 × 10−4 mol
So, millimoles = 15.71 × 10−4 mol ×1000 = 1.571 mmol