125pex2sol

# 125pex2sol - Copyright RJZ 1 no unauthorized use allowed...

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Copyright RJZ 2/17/12 1 no unauthorized use allowed Chemistry 125 - W12 Solutions for Practice Midterm 2 This material is copyrighted. Any use or reproduction is not allowed except with the expressed written permission of Dr. Zellmer. If you are taking Chem 125 you are allowed to print one copy for your own use during the quarter you are taking Chem 125 with Dr. Zellmer. You are not allowed to disseminate this material to anyone else during the quarter or in the future.

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Copyright RJZ 2/17/12 2 no unauthorized use allowed 1) Freezing point depression problem. ) T f = i K f m = m part K f K f = freezing point depression constant (depends on solvent) i = # ions in the formula for an ionic cmpd or ionizing molecule = 1 for nonionizing or nondissociating cmpds m = molality ( i m = m part , molality of particles) for an “ideal” ionic solution, i = # ions from formula for the compound H 2 O Na 3 PO 4 (s) --------> 3 Na + + PO 4 3 ! i = 4 ions (PO 4 3 ! stays together as a single particle in soln) Need molality. Given mass so need to get moles first and then molality. 1 mol Na 3 PO 4 mol Na 3 PO 4 = 0.3279 g Na 3 PO 4 x ----------------------- = 0.00200 0 1 mol Na 3 PO 4 163.94 g Na 3 PO 4 0.00200 0 1 mol Na 3 PO 4 molality of Na 3 PO 4 = ------------------------------ = 0.0200 0 1 mol Na 3 PO 4 0.1000 kg H 2 O ) T f = 4 (1.86°C/ m ) (0.0200 0 1 m ) = 0.14 8 8°C = 0.14 9 °C (3 s.f.) The freezing point of a solution at 1 atm is less than that of the pure solvent by ) T f , T f_soln = T f_solvent - ) T f = 0 °C - 0.149 °C = ! 0.14 9 °C The freezing point of an aqueous solution has to be negative since it will be less than 0 °C. C
Copyright RJZ 2/17/12 3 no unauthorized use allowed 2) Vapor pressure of a volatile substance is given by Raoult’s Law and the vapor pressure lowering due to a nonvolatile solute can be derived from this. P A = O A P A 0 Raoult’s Law A = mole fraction of solvent (CH 3 OH) in the solution P A 0 = vapor pressure of pure solvent P A = vapor pressure due to the solvent. This is also the vapor pressure of the solution since the solvent is the only thing contributing to the vapor pressure of the solution. For a nonvolatile solute the vapor pressure lowering due to addition of solute, ) P, is given by: ) P = B P A 0 B = mole fraction of solute in the solution ) P = P A 0 - P A ) P = 160 torr - 120 torr = 4 0 torr glycerol = ) P/P A 0 = (40 torr)/(160 torr) = 0.25 0 0 or P soln = P CH3OH = CH3OH P 0 CH3OH CH3OH = P soln /P 0 CH3OH = (120 torr)/(160 torr) = 0.750 glycerol = 1 - CH3OH = 1 - 0.750 = 0.250 C

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Copyright RJZ 2/17/12 4 no unauthorized use allowed 3) Want the mass of sucrose, C 12 H 22 O 11 , that needs to be dissolved in 4.00 x 10 2 mL of solution to produce a solution with an osmotic pressure of 494 torr at 25.0°C. To find the mass of sucrose you use the osmotic pressure equation to determine molarity (or moles directly). . J = i M R T , where J = osmotic pressure i = # ions in the formula for an ionic cmpd or ionizing molecule = 1 for nonionizing or nondissociating cmpds J = M part R T M part = molarity of particles ( i M = M part , molarity of particles) R = 0.0821L C atm/mol C K; T = temp in kelvin J V = n R T ( i = 1) Since C 12 H 22 O 11 is nonionizing i = 1.
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125pex2sol - Copyright RJZ 1 no unauthorized use allowed...

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