Chap07 soln

# Physical Chemistry

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7 Simple mixtures Solutions to exercises Discussion questions E7.1(b) For a component in an ideal solution, Raoult’s law is: p = xp . For real solutions, the activity, a , replaces the mole fraction, x , and Raoult’s law becomes p = ap . E7.2(b) All the colligative properties are a result of the lowering of the chemical potential of the solvent due to the presence of the solute. This reduction takes the form µ A = µ A + RT ln x A or µ A = µ A + RT ln a A , depending on whether or not the solution can be considered ideal. The lowering of the chemical potential results in a freezing point depression and a boiling point elevation as illustrated in Fig. 7.20 of the text. Both of these effects can be explained by the lowering of the vapour pressure of the solvent in solution due to the presence of the solute. The solute molecules get in the way of the solvent molecules, reducing their escaping tendency. E7.3(b) The activity of a solute is that property which determines how the chemical potential of the solute varies from its value in a specified reference state. This is seen from the relation µ = µ + RT ln a , where µ is the value of the chemical potential in the reference state. The reference state is either the hypothetical state where the pure solute obeys Henry’s law (if the solute is volatile) or the hypothetical state where the solute at unit molality obeys Henry’s law (if the solute is involatile). The activity of the solute can then be defined as that physical property which makes the above relation true. It can be interpreted as an effective concentration. Numerical exercises E7.4(b) Total volume V = n A V A + n B V B = n(x A V A + x B V B ) Total mass m = n A M A + n B M B = n(x A M A + ( 1 x A )M B ) where n = n A + n B m x A M A + ( 1 x A )M B = n n = 1 . 000 kg ( 10 3 g / kg ) ( 0 . 3713 ) × ( 241 . 1 g / mol ) + ( 1 0 . 3713 ) × ( 198 . 2 g / mol ) = 4 . 670 ¯ 1 mol V = n(x A V A + x B V B ) = ( 4 . 670 ¯ 1 mol ) × [ ( 0 . 3713 ) × ( 188 . 2 cm 3 mol 1 ) + ( 1 0 . 3713 ) × ( 176 . 14 cm 3 mol 1 ) ] = 843 . 5 cm 3 E7.5(b) Let A denote water and B ethanol. The total volume of the solution is V = n A V A + n B V B We know V B ; we need to determine n A and n B in order to solve for V A . Assume we have 100 cm 3 of solution; then the mass is m = ρV = ( 0 . 9687 g cm 3 ) × ( 100 cm 3 ) = 96 . 87 g of which ( 0 . 20 ) × ( 96 . 87 g ) = 19 . 374 g is ethanol and ( 0 . 80 ) × ( 96 . 87 g ) = 77 . 496 g is water. n A = 77 . 496 g 18 . 02 g mol 1 = 4 . 3 0 mol H 2 O n B = 19 . 374 g 46 . 07 g mol 1 = 0 . 42 05 mol ethanol

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98 INSTRUCTOR S MANUAL V n B V B n A = V A = 100 cm 3 ( 0 . 42 05 mol ) × ( 52 . 2 cm 3 mol 1 ) 4 . 3 ¯ 0 mol = 18 . 15 cm 3 = 18 cm 3 E7.6(b) Check that p B /x B = a constant (K B ) x B 0 . 010 0 . 015 0 . 020 (p B /x B )/ kPa 8 . 2 × 10 3 8 . 1 × 10 3 8 . 3 × 10 3 K B = p/x , average value is 8 . 2 × 10 3 kPa E7.7(b) In exercise 7.6(b), the Henry’s law constant was determined for concentrations expressed in mole fractions. Thus the concentration in molality must be converted to mole fraction.
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