Chemistry, asked by emmasam, 10 months ago

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

Answered by deerajdr
0

Answer:

this may help you

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

Answered by Anonymous
3

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

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