Chap07 soln

Physical Chemistry

Info iconThis 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 + 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 speci±ed reference state. This is seen from the relation µ = µ + 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 de±ned 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 . 1g / mol ) + ( 1 0 . 3713 ) × ( 198 . 2g / mol ) = 4 . 670 ¯ 1 mol V = n(x A V A + x B V B ) = ( 4 . 670 ¯ 1 mol ) × [ ( 0 . 3713 ) × ( 188 . 2cm 3 mol 1 ) + ( 1 0 . 3713 ) × ( 176 . 14 cm 3 mol 1 ) ] = 843 . 5cm 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
Background image of page 1

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

View Full DocumentRight Arrow Icon
98 INSTRUCTOR S MANUAL V n B V B n A = V A = 100 cm 3 ( 0 . 42 05 mol ) × ( 52 . 2cm 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.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 01/29/2008.

Page1 / 15

Chap07 soln - 7 Simple mixtures Solutions to exercises...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online