Calculate the mole fraction of ammonia in water at an ammonia partial pressure of 120. torr.
Hey there @whitneym9
If we use Henry's Law equation we can get the molarity of ammonia.
120 torr x (1 atm/ 760 torr) = 0.1579 atm
Sgas = kH x Pgas = (2.700 x 10 mol L^-1 atm^-1) (0.1579 atm)
At this point we would also need the density of the solution in order to determine the mole fraction of ammonia.
Let's say for example we determined the molarity as being 4.263 M and the density of the solution as being 1.310 g/mL. In order to determine the mole fraction we would do the following:
4.263 M NH3 ---> 4.263 moles NH3 / 1 L of solution
1 L = 1000 mL
1000 mL x 1.310 g / 1 mL = 1310 g solution
4.263 moles NH3 x (17.034 g NH3 / 1 mole NH3) = 72.62 g NH3
1310 g solution - 72.62 g NH3 = 1237.38 g H2O
Now in order to determine the mole fraction we would need to change the grams into moles
Mole fraction (X) of NH3 = [ (4.263 moles NH3) / (4.263 moles NH3 + 68.68 moles H2O) = 0.058
Hope that helps. Please let us know if you still have a question in terms of this problem.
How did you determine the density of 1.310g?