Question

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Here's a question for you☺️❤️
Calculate the solubility of H2 in water at 25°C if it's partial pressure above the solution is 1 bar. Given that KH for H2 in water at 25℃ is 71.18 kbar.
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Answer 1

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PH2=KHXH2

XH2=PH2kH

=1bar7.18×103bar

=1.404×10−5

Now,

nH2O in 1L=1000g18

=55.5

Therefore ,

nH255.55=1.404×10−5

nH2=1.79×10−4moles

Hence the solubility of H2 in water is 7.79×10−4mol/L

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Answer 2

Explanation:

Heya mate

The answer is

Temp = 25°C

P( H2) =KHXH2

=1bar7.18×103bar

=1.404×10 ^-5

nH2O in 1L= 1000g

=55.5

Then

nH255.55=1.404×10−5

nH2=1.79×10−4 moles

Hence the solubility of H2 in water is 7.79×10−4mol/L

hope it helps