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Ch14_Solutions_3

# Ch14_Solutions_3 - Pressure Calculations for Solution of 2...

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Pressure Calculations for Solution of 2 Volatile Liquids Dalton’s Law of Partial Pressures P total = P A + P B Use Raoult’s Law to calculate partial pressure above each volatile solvent P A = X A P A ° P B = X B P B ° Combine to get total pressure above solution P total =X A P A ° + X B P B °

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Ex. Benzene and Toluene Consider a mixture of benzene, C 6 H 6 , and toluene, C 7 H 8 , containing 1.0 mol benzene and 2.0 mol toluene. At 20 °C, the vapor pressures of the pure substances are: benzene = 75 torr toluene = 22 torr Assuming the mixture obeys Raoult’s law, what is the total pressure above this solution?
Ex. Benzene & Toluene (cont.) 1. Calculate mole fractions of A & B. 1. Calculate partial pressures of A & B 1. Calculate total pressure torr torr P X P benzene benzene benzene 25 75 33 . 0 = × = = ( 29 benzene   33 . 0 0 . 2 0 . 1 0 . 1 = + = mol mol X benzene ( 29 toluene   67 . 0 0 . 2 0 . 1 0 . 2 = + = mol mol X toluene torr torr P X P toluene toluene toluene 15 22 67 . 0 = × = = torr torr P P P toluene benzene total 40 ) 15 25 ( = + = + =

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Learning Check The vapor pressure of 2-methylheptane is 233.95 torr at 55°C. 3-ethylpentane has a vapor pressure of 207.68 at the same temperature. What would be the pressure of the mixture of 78.0 g 2- methylheptane and 15 g 3-ethylpentane? ( 29 ( 29 torr   207.68 0.17 torr   233.95 0.82 P 26 737 × + × = P solution = X A P o A + X B P o B = 230 torr mol   682 0 g/mol   23 114 g   0 78 ane methylhept - 2   mole 83 . . . = = mol   14 0 g/mol   00.2 1 g   5 1 ne ethylpenta - 3   mole 97 . = = X 2-methylheptane = 0.682 83 /(0.682 83  + 0.14 92 )= 0.82 737 X 2-methylheptane  = 0.14 92 /(0.682 83  + 0.14 92 )= 0.17 26
Solutes also Affect Freezing & Boiling Points of Solutions Facts : Freezing Point of solution always Lower than pure solvent Boiling Point of solution always Higher than pure solvent Why? Consider the phase diagram of H 2 O Solid, liquid, gas phases in equilibrium Blue lines P vs T

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Pure Water Triple Point All 3 phases exist in equilibrium simultaneously Pure H 2 O Dashed lines at 760 torr (1 atm) that intersect solid/liquid & liquid/gas curves Give T for Freezing Point (FP) & Boiling Point (BP) T FP T BP
Solution—Effect of Solute Solute molecules stay in solution only None in vapor None in solid Crystal structure prevents from entering Liquid/vapor # solvent molecules entering vapor Must go to higher T to get all liquid to gas Line moves to higher T all along phase border ( red )

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Solution—Effect of Solute Triple point lower & to left ( ) Solid/liquid Solid/liquid line to left ( red ) Lower T all along phase boundary Solute keeps solvent in solution longer Must go to lower T to form crystal
Freezing Point Depression & Boiling Point Elevation In solution , observe BP & FP over pure solvent Presence of solute, depresses FP & elevates BP Both T f & T b depend on relative amounts of solvent & solute Colligative properties boiling point elevation Increase in boiling point of solution vs. pure solvent freezing point depression Decrease in freezing point of solution vs. pure solvent

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