Question

what partial pressure of He gas (in mm Hg) is required to maintain a solubility of 3.43x10^-4 g/L in water at 25 degrees C?

kH for He at 25 degrees C is 3.26x10^-4 mol/L•atm.

_________mm Hg

Answer 1

Ans :

Solubility of He = 3.43 x 10^-4 g/L

molar mass of He = 4.0026 g/mol

So solubility of He in M = 3.43 x 10^-4 g/L / 4.0026 g/mol

= 8.6 x 10^-5 M

Now :

C = kH x P

putting values :

8.6 x 10^-5 M = 3.26 x 10^-4 M/atm x P

P = 0.263 atm

partial pressure in mmHg = 0.263 x 760 mmHg

= 200 mmHg