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Unformatted text preview: Chemistry 132, Winter 2008 Solutions to Homework No. 2 Problem 1. See Levine problem 12.1. (a) True. Addition of a solute B at constant T and P to pure solvent A decreases the mole fraction x A of the solvent. Since ∂μ A ∂x A T,P,n j 6 = A > 0 (Levine eq. 4.90), then addition of the solute lowers μ A . (b) True. Adding solute decreases x A , which decreases μ A as in part (a). Problem 4. See Levine problem 12.15. Let A , C , D , and B represent benzene, naphthalene, anthracene, and the mixture of C + D . We want to find n C and n D . m B = n C + n D w A (1) ⇒ n C = m B w A n D , (2) w B = n C M C + n D M D (3) = ( m B w A n D ) M C + n D M D , (4) ⇒ n D ( M D M C ) = w B m B M C w A (5) ⇒ n D = w B m B M C w A M D M C (6) = w B + Δ T f M C w A /k f M D M C , (7) where Levine eq. 12.15 was used to obtain equation 7. Us ing the known values, w B = 6 . 0 g, w A = 300 g, Δ T f = . 70 ◦ C, k f = 5 . 1 ◦ kg mol 1 , M C = 128 . 17352 g / mol, M D = 178 . 2334 g / mol, in equations 7 and 2 gives n D = . 01443 mol and n C = 0 . 02675 mol. Problem 7. See Levine problem 12.26. In part (b) use m B, 1 = . 012 mol / kg and m B, 2 = 0 . 120 mol / kg. (a) Apply Levine eq. 12.20 to the solutions in question and follow the same derivation given on page 347: μ A, 1 = μ A, 2 (8) μ * A ( P,T ) + RT ln γ A x A, 1 (9) = μ * A ( P + Π ,T ) + RT ln γ A x A, 2 (10) μ * A ( P + Π ,T ) μ * A ( P,T ) (11) = Z P +Π P V * m ,A dP (12) = Π V * m ,A (13) ⇒ RT ln γ A x A, 2 RT ln γ A x A, 1 = Π V * m ,A . (14) For ideally dilute solutions ln γ A x A ≈  x B (Levine eq. 12.11), so equation 14 becomes RT ( x B, 2 + x B, 1 ) = Π V * m ,A (15) ⇒ Π = RT V * m ,A ( x B, 2 x B, 1 ) . (16) When solution 1 is pure solvent, x A, 1 = 1 and x B, 1 = 0, so equation 16 becomes the same as Levine eq. 12.25. (b) First express mole fractions in terms of molalities: m B = n B w A = n B n A M A (17) ⇒ n A n B = 1 m B M A , (18) x B = n B n B + n A (19) = 1 n A /n B + 1 (20) = 1 1 /m B M A + 1 . (21) Using the given solute molalities, m B, 1 = 0 . 021 mol / kg and m B, 2 = 0 . 120 mol / kg, together with M A = 18 . 01528 g / mol in equation 21 yields x B, 1 = 3 . 78178 × 10 4 and x B, 2 = 2 . 15717 × 10 3 . With these mole fractions and with T = 298 . 15 K, R = 8 . 3145 JK 1 mol 1 , and V * m ,A = M A /ρ A = (18 . 01528 g / mol) / (10 6 g / m 3 ), equation 16 gives Π = 244796Pa = 2 . 416 atm as the osmotic pressure at equi librium....
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This note was uploaded on 05/22/2008 for the course CHEM 132 taught by Professor Lindenberg during the Winter '08 term at UCSD.
 Winter '08
 Lindenberg
 Mole

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