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CH131_10B_11A

CH131_10B_11A - A Plot of P vs T B Shows regions of 3...

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A. Plot of P vs. T B. Shows regions of 3 phases as a function of T & P, & regions of phase equilibria 1. points. x, y, z in gas, liq. solid regions for H 2 O 2. Lines indicate T, P conditions where 2 phases coexist a. TC => l g coexist equil (our v.p. vs T from before) b. AT => s g coexit equil c. TB => s coexist equil 3. Point T => 3 equil lines intersect; all 3 phases coexist => Triple point 4. Point C => critical point ; P&T above which liq-vap coexist curve ends; supercritical fluid region x z y Pressure (atm)
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Pressure (atm) Above, P c (218 atm) & T c (374.2 o C), critical pressure & critical temperature, one phase only => can’t distinguish between liq.& vap; have a fluid described as dense gas or dilute liquid; super critical fluid (SCF) above P c , T c (single phase) P c, T c Consider the H 2 O diagram start @ P = 1atm, and add heat => T increases until s interface; heat then goes into melting solid; added heat raises T liq. until 100 o C, b.p., l g equil. then just gas T >T b If we started at P < 0.006 atm (< triple pt.) and heat, solid gas directly without appearance of liq.
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Chapter 11 Read 11.1, 11.5-11.6 Problems Chap. 11 - 1, 3, 5, 11, 41, 43, 45, 47, 51, 53, 55, 57, 59, 61
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Chap. 11 Solutions Just finished describing some properties of pure substances in gas, liquid or solid phases. Many chemical systems are mixtures - in particular, let’s consider solutions => homogeneous mixtures of 2 (or more) substances Solvent => corresponds to the major component of solution Solute => minor component(s) of solution Several ways to quantitatively characterize the composition of solutions…..
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1. Mole fraction (already discussed in context of Dalton’s law and partial pressures) Remember: mole fraction of component 1 x 1 = n 1 n Total In a 2-component system (1, 2) n Total = n 1 + n 2 , so; x 1 = n 1 n 1 + n 2 x 2 = n 2 n 1 + n 2 x 1 + x 2 =1 Thus, x 2 =1- x 1 2. Molarity (M) a very common description of solution composition Molarity = # moles solute liter of solution A convenient description of concentration => only problem, … M is T dependent (volume is T-dependent)
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3. Molality(m) molality = # moles solute kg solvent = mol kg - 1 ( m ) m is temperature independent. At low concentrations in aqueous solutions m M Because d H2O ~ 1 g/ml ~1 kg/L so m = mol kg solvent 1 kg L mol L of solution = M When d not ~ 1 g/ml (nonaq. solvent or concentrated solutions) m M
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Example of use of solution units… 2.45 M solution of aq. methanol (CH 3 OH) has d = 0.976 g/ml (1ml = 1 cm 3 = 1 cc) What is the molality of this solution? molality = m = # moles solute kg solvent From definition of M, 1 L of this solution => 2.45 mol CH 3 OH (solute) How many kg H 2 O (solvent) in 1 L of this solution? 1 L solution 0.976 g / ml 10 3 ml /1 L = 976 g solution 976 g due to weight of both solute (CH 3 OH) and solvent (H 2 O) but 2.45 mol CH 3 OH x 32.04 g/mol = 78.9 g CH 3 OH # g H 2 O in 1 L solution = 976 g – 78.9 g = 898 g H 2 O, so m = 2.45 mol CH 3 OH 0.898 kg H 2 O = 2.73 m
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Colligative Properties of Solutions These are properties of solutions that just depend on the
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