95 
A review problem in which one kind of concentration unit is converted into several others. (This material was covered in General Chemistry.) 
97 
Review problem in which the solution density is used to convert between molarity and molality. (This material was covered in General Chemistry.) 


934 
A T/F question. Part (e) is optional, although not really difficult. Part (b) is especially important. Reactions and processes like dissolution only happen if the free energy decreases as a result of the reaction or process. If free energy does not decrease then the process (in this case formation of a solution) doesn’t happen. It is necessary to read part (d) very carefully. For part (f) it is necessary to know that the partial molal volume is for component i is (¶V/¶n_{i}), where T, P, and all the other n_{j} are kept constant. If the solution is ideal, which means that the volumes of the components add to give the solution volume, the partial molal volume does not depend on the number of moles present of any other component. 
936 
The first of three problems that are based on the same solution, which is to be considered ideal. (If the solution were not ideal there would not be enough information to do the problem). In this problem thermodynamic quantities are to be calculated. The formula for calculating the entropy of mixing is the same in a solution as it is in the gas phase (D_{mix}S = R S n_{i}lnX_{i} = N_{tot}R S X_{i}lnX_{i}) because in both the mixing is at the molecular level. 
937 
(Second problem in the series). Exercise based on Raoult's Law. The mole fractions in the gas (or, vapor) above the solution are related (by Dalton’s Law) to the partial pressures: X_{i}^{vap} = Y_{i} = P_{i}/P_{tot}. (The convention used by chemical engineers, who use X_{i} for X_{i}^{liq} and Y_{i} for X_{i}^{vap} is quite convenient.) 
941 
(Third problem in the series). Density of an ideal solution. For an ideal solution D_{mix}V^{o}= 0, which means that the liquid volumes are additive (50 mL of A + 50 mL of B gives 100 mL of solution). The masses of the liquids are given, but since density is an intensive property the actual masses of the liquids are unimportant but it does matter that they are equal. (Please remember that adding or subtracting intensive properties doesn’t work; the answer obtained is almost certain to be wrong. Only extensive properties can be added or subtracted). 
932 
A T/F question. Part (c) is especially important. A component cannot go into solution unless the process of dissolution corresponds to a lowering of the free energy. 
948 
Vapor pressures, etc. for a solution that is ideally dilute. In an ideally dilute solution the solvent (X_{A} near 1) follows Raoult's Law and the solute (X_{B} small) follows Henry's Law. Levine always gives Henry’s Law in the form P_{B} = X_{B}K_{B} although some books use other forms. 


954 
A straightforward problem involving Henry's Law. The two gases behave independently. The solubility of gases in liquids is limited. The gas solubility is described by Henry’s Law with X_{gas} = P_{gas}/K_{gas}, where K_{gas} is the Henry’s Law constant for the gas. 
962 
An exercise combining Trouton's rule (use the form D_{vap}S = 10.5 R), Raoult's law, and the ClausiusClapeyron equation. This calculation is longer than most. 
968 
A thought problem that points out differences between ideal solutions and ideally dilute solutions. In an ideally dilute solution it is still true that m_{B} = m_{B}° + RTlnX_{B}, but m_{B}° (which will be discussed later) is not the molar free energy of the pure substance. In part (d) the V_{i} with a bar over it is the partial molar volume, (∂V/∂n_{i})_{T,P,nj}, which is analogous to m, which is the partial molar free energy. (The equation in part (d) was written on the board in connection with the derivation of the GibbsDuhem equation). Please ignore part (h), to which no answer is given in the answer book. 