Hey guys, so I did my Kaplan FL #5 yesterday and just been going over the PS section and the very first question is related to the thread title. The passage talks about how acids dissociate more in ammonia than in water since ammonia acts as a base. Then they say that the student prepares these two solutions: Solution X: 2 mols of CH3COOH dissolved in 2000g of ammonia (liq) Solution y: 2 mols of CH3COOH dissolved in 2000g of water [Kf_amm = -0.957 C/m ; Kf_water = -1.86 C/m] The question then asks whether the relative freezing point depression of X or Y is larger. The answer to the question was Solution X. Their explanation was that the acid dissociates more readily in ammonia than in water. They talk about how the dissociation constant of the acid is about 10^-5 in water (wasnt given in passage but roughly) while in ammonia, it would be full dissociation. They say that this factor is much more significant than the difference in kf's. I don't understand this explanation.... We know that the magnitude of kf of water is double that of ammonia. However, I don't understand how the dissociating factor is so much more significant. Lets say that none of the acid dissociated in water so the value of i in [deltaT = iKm] is 1 (the acid is still dissolved in water right?), and in full dissociation in ammonia, i = 2. So if we do the math here: kf_water = 2 x kf_ammonia 2 x i_water = i_ammonia m_water = m_ammonia So the 2's both cancel out and the depression should be around equal? Please tell me if I am wrong... This is frustrating me.