Helium is dissolved in water with the total pressure of gas (Helium + water vapor) over the water being 1.60

Question

Helium is dissolved in water with the total pressure of gas (Helium + water vapor) over the water being 1.60 atm. The temperature is 25o C. What is the concentration of Helium in the water?
Give your answer in g/L .
The vapor pressure of water at 25.0oC is 23.8 mm Hg.


Note that kH has units of mol/L / mmHg

Answer 1

ok to do it you take the pressure of He 1.60atm and x by 760 mm Hg to convert = ____ mmHg - 23.8 (vapor press of water). then multiply by henry's constant for He (3.7e-4). take that ans which will be in mol of He/ L and multiply by the molar mass of He (4g) and u will get ur ans in g/L...

the ans should be: 1.764456
Good luck! :)

Answer 2

You must have the same teacher as another girl who posted here, though she had a different pressure 2.2 vs 1.6.

I think Henry’s Law is what you are looking for

http://en.wikipedia.org/wiki/Henry%27s_Law

In this page they list the Henry’s K value for Helium at 3.7 x 10-4 mol / (L* ATM).

Converting your vapor pressure 23.8 mm HG * .001316 = 0.0313 ATM

Could your answer be as simple as (3.7 x 10 –4) * (1.60 – .0313) = 0.580 mM/L

Answer 3

After seeing the answer, pls read Psalms 23:1.

This is the answer:
PV=nRT

P=pressure
V=volume
n=moles
R=universal gas constant
T=temperature

Ptotal = PHe + Pwater

Solution:
Assume: the gases behave ideally

1.6 = PHe + 23.8(1/760)
PHe = 1.57 atm
From IGE,
(1.57)V = n(0.0821)(298)
V/n = 15.58 mol/L
= 31.17 g/L

Answer 4

mopet mopet