Chap07 soln

Physical Chemistry

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern