Asked by Daiyo
Calculate the aqueous solubility (in milligrams per liter) of the following. The temperature is 20.°C in each case, the pressures are partial pressures of the gases.
(a) air at 2.4 atm
(b) He at 1.7 atm
(c) He at 11 kPa
There's also a table giving Henry's Constants for air and helium and they are 7.9 x 10^-4 and 3.7 x 10^-4 respectively.
(a) air at 2.4 atm
(b) He at 1.7 atm
(c) He at 11 kPa
There's also a table giving Henry's Constants for air and helium and they are 7.9 x 10^-4 and 3.7 x 10^-4 respectively.
Answers
Answered by
Daiyo
Oh and the given molar mass of air was given to be 28.97g/mol
Answered by
DrBob222
I looked here on the internet to see what form of Henry's Law you were/are using. Here is a link.
http://en.wikipedia.org/wiki/Henry%27s_law
It appears you are using pk = c with p in atm, and c in moles/L.
a)So pk = c
p = 2.4 atm and k = 7.9E-4
Substitute and solve for c. I find about 0.00190 M or 0.00190 moles/L.
moles = grams/molar mass; therefore, grams air = moles x molar mass air and that converted to mg should do it.
The others are done the same way.
http://en.wikipedia.org/wiki/Henry%27s_law
It appears you are using pk = c with p in atm, and c in moles/L.
a)So pk = c
p = 2.4 atm and k = 7.9E-4
Substitute and solve for c. I find about 0.00190 M or 0.00190 moles/L.
moles = grams/molar mass; therefore, grams air = moles x molar mass air and that converted to mg should do it.
The others are done the same way.
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