A handbook lists the value of the Henry's Law constant as 2.700 x 10 mol L-1 atm-1 for ammonia, NH3, dissolved in water at 25 C.

asked by @whitneym9 • about 1 year ago • Chemistry • 5 pts

Calculate the mole fraction of ammonia in water at an ammonia partial pressure of 120. torr.

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2 answers

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.

answered by @jules • about 1 year ago

How did you determine the density of 1.310g?

answered by @shanar12 • 3 months ago