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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 133 �ese two expressions for ptot are set equal and the resulting equation rear- ranged to �nd xA p∗AxA yA = p ∗ B(1 − xA) 1 − yA hence xA = p∗B yA p∗A(1 − yA) + p∗B yA With the data given xA = p∗B yA p∗A(1 − yA) + p∗B yA = (82.1 kPa) × (0.612) (68.8 kPa)∗(1 − 0.612) + (82.1 kPa) × (0.612) = 0.653... and xB = 1 − 0.653... = 0.346... �e composition of the liquid is therefore xA = 0.653 and xB = 0.347 . �e total pressure is computed from pA = ptot yA and pA = p∗AxA to give ptot = xAp∗A/yA ptot = xAp∗A yA = (0.653...) × (68.8 kPa) 0.612 = 73.4 kPa E5B.12(b) If the solution is ideal, the vapour pressure of component J in the solution obeys Raoult’s law, [5A.22–151], pJ = p∗J xJ, where xJ is the mole fraction in the solution. In the gas the partial pressure is pJ = yJptot, where yJ is the mole fraction in the vapour. Assuming ideality, the total pressure is computed as ptot = pA + pB = p∗AxA + p∗B(1 − xA) = (110.1 kPa) × (0.4217) + (76.5 kPa) × (1 − 0.4217) = 90.7 kPa �e normal boiling point is when the total pressure is 1 atm, but the pressure predicted by Raoult’s law is signi�cantly di�erent from this. Raoult’s law there- fore does not apply and the solution is not ideal . It is not possible to �nd yA, the composition of the vapour, from pA = ptot yA because pA cannot be computed from the data given. Solutions to problems P5B.2 �e freezing point depression ∆Tf is related to the molality of the solute B, bB, by [5B.12–161], ∆Tf = KfbB, where Kf is the freezing-point constant. �e molality corresponding to the given freezing-point depression is therefore bB,app = ∆Tf Kf = 0.0703 K 1.86 K kg mol−1 = 0.0377... mol kg−1 �e stated molality of the solution is 0.0096 mol kg−1, so it is evident that there are more species in the solution than the number of molecules of�(NO3)4 added. �e ratio of bB,app to bB gives an indication of the number of species present per molecule dissolved bB,app bB = 0.0377... mol kg −1 0.0096 mol kg−1 = 3.93...