Physical Chemistry , 1st ed.

Solution Refer to Figure 7.17. Using the tie lines to connect the vapor composition for each liquid phase composition, we ultima ...

liquids are immiscible, so the total pressure in equilibrium with the liquids is simply the sum of the two equilibrium vapor pre ...

The way to make a proportionality an equality is to define a proportionality constant Ki, so now we have piKixi (7.31) where th ...

7.6 Liquid/Solid Solutions In this section, we will consider only solutions in which the liquid component has the majority mole ...

solubility is (grams of solute)/(100 mL of solvent).] Most of the solutions we work with are unsaturated, having less than the m ...

point, (^) fusGMPequals zero, so we are simply adding zero to equation 7.38. We g e t ln xdissolved solute R fu T sG M M P P ...

experimental values, especially considering the assumptions made in deriv- ing equation 7.39. Example 7.11 Use equation 7.39 to ...

Solid solutions should be distinguished from composites,which are mate- rials formed from two or more solid components that neve ...

example, we can start with two pure components A and B, which have specific melting points (MPs), as shown in the temperature-co ...

A begins to melt. This reduces the amount of A in the solid (indicated by the dotted line above the solid line in Figure 7.22b). ...

One important application of the detailed understanding of solid solution phases is called zone refining,which is a method for p ...

7.8 Colligative Properties Consider the solvent of a solution. It is typically defined as the component with the majority mole f ...

in molarity units varies due to expansion or contraction of the solution’s volume. The next colligative property is boiling poin ...

in temperature of the equilibrium melting or freezing process. Equation 7.44 becomes xsolute R T fu 2 M sH P  Tf (7.45) The r ...

expression, this implies that the unit molality can be substituted in the de- nominator. Therefore, the final answer is Kf20.83 ...

different pressure than the column on the right side. The difference in the two pressures, represented by the difference in colu ...

Again, consider that ln xsolvent,solutionln(1 xsolute) xsolute. Making one final substitution: xsoluteRTV This is usuall ...

Using the equation Fma, this force corresponds to a mass of 248 Nm9.81  m s^2  m25.3 kg where we have used the fact that 1 ...

Although 0.010 molal is not a very concentrated solution, the predicted osmotic pressure effects are substantial. Osmotic pressu ...

Using the fact that 0.0100 g was used to make 1.00 L of solution, we have the relationship 0.000123  m L ol 0.0100  L g  Th ...