The solubility of NH3 in H20 is 518 g/L at a partial pressure of 760.0 torr.
1. Calculate the value of the Henry's Law constant for NH3.
2. calculate the solubility of NH3 in H2O at a partial pressure of 225.0 torr.
2 answers
The subject is not Pensacola State
I worked this for JD. Assuming you are not JD, here is what I wrote (more or less).
There are all kinds of unts used for Henry's Law constant and different equations for Henry's Law. The one I use is p = kc where p is in atm and c is in moles/L. That gives k = p/c = atm*L/moles for k. Convert 518 g NH3 to moles. moles = grams/molar mass and that is in 1 L; therefore, that is the molarity. 518/17 = 30.47 and k = 1/30.47. For the remaining part of the problem, p = kc = (225/760) = c/30.47 and solve for c.
p = kc and (225/760)
There are all kinds of unts used for Henry's Law constant and different equations for Henry's Law. The one I use is p = kc where p is in atm and c is in moles/L. That gives k = p/c = atm*L/moles for k. Convert 518 g NH3 to moles. moles = grams/molar mass and that is in 1 L; therefore, that is the molarity. 518/17 = 30.47 and k = 1/30.47. For the remaining part of the problem, p = kc = (225/760) = c/30.47 and solve for c.
p = kc and (225/760)
2 answers