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Unformatted text preview: Answers to 12 and 13 1. ΔT = T f T i = K f im solute Solute = 20 g Benzene = 500 g T f = 3.77 T i = 5.48 K f = 5.12 (3.77  5.48) = 5.12 x 1 x ((20g/MW)/500 g) x 1000 g/kg MW = 119.8 g, A 2. Henry’s Law: X A =kP A mol CO 2 /(mol CO 2 + mol H 2 O) = 6.1 x 104 x 1.1 Y/(Y + (0.50L x 1000 g/L x (1 mol/18.02 g))) = (6.1 x 104 x 1.1) Y = 0.0186 mol CO 2 Based on the very small constant, the above calculation assumed that the moles of CO2 dissolved in the club soda was insignificant, i.e., that all of the 0.50 L of club soda is water. Proof that this assumption is valid is that 0.0186 mol CO 2 is only 0.07% of the 0.50 L of H 2 O. However, it wasn’t necessary to make this assumption. In the calculation the moles of CO 2 could have been subtracted from the moles of H 2 O : ((Y)/(Y + (0.50 x 1000 x (1/18.02) – Y))) = (6.1 x 104 x 1.1) Y = 0.0186 mol CO 2 3. Raoult’s Law: P soln = X solvent P o solvent mol H 2 O = 200g/18.02gmol1 = 11.099 mol P soln = ((11.099 mol H 2 O)/((11.099 mol H 2 O + 0.300 molsucros))) x 355 Torr = 345.7 Torr 4. This is a more difficult problem than normally seen with Raoult’s Law. Psolv = X solv x P o solv Benzene Toluene Pure 95.1 torr Pure 28.4 torr 1.0 mol 2.0 mol P solv = (1.0/3.0) x 95.1 2/3 x 28.4 = 31.70 vapor pressure = 18.93 vapor pressure 31.70/(31.70 + 18.93) = 0.626 5. E A. Absence of significantly favorable solutesolute interactions is saying the same thing as the presence of significantly unfavorable solutesolute interactions. The presence of significantly unfavorable solutesolute interactions never helps miscibility (i.e., dissolution); if anything, if the solutesolute interaction is so unfavorable, i.e., if the ΔH is dissolution); if anything, if the solutesolute interaction is so unfavorable, i....
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This note was uploaded on 04/05/2008 for the course CHEM 161, 101 taught by Professor Hansell during the Spring '08 term at Rutgers.
 Spring '08
 Hansell

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