Key Exam 2_SU_07 - KEY and Solutions Guide Exam 2 C112 Summer 2007 1(a wrong The vapor pressure of the solvent increases as its mole fraction

Key Exam 2_SU_07 - KEY and Solutions Guide Exam 2 C112...

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KEY and Solutions Guide Exam 2 C112 Summer 2007 1. (a) wrong. The vapor pressure of the solvent increases as its mole fraction increases. (as the mole fraction of all solutes decreases) (b) correct. All gases have increased solubility at lower temps. (c) correct . Pure solvent has a mole fraction of solvent, X solvent = 1.0. This is the condition that ensures the highest vapor pressure of the solvent. All surface sites on the solvent are occupied by solvent molecules, increasing the probability that they will escape into the atmosphere above. (d) correct . This is a verbal way of stating Henry’s law. S gas = K H P gas (e) correct . The number of “disruption sites” in the solvent “lattice” is the basis of the colligative effects. The formula molality of the solute is multiplied by the average number of solution phase particles it generates, to get a true measure of its impact on colligative properties. 2. Strategy: Lowest vapor pressure will correspond to greatest number of disruption sites in the solvent “lattice”. Highest v.p occurs for the pure solvent. Examine the solutes to see if one has a higher concentration of solute. If a mixture is used, the concentrations are added. None of the solutions is higher than another, all are 0.2 m, except for the pure solvent so answer (e) is eliminated. Next, determine the number of dissolved “pieces” formed by each solute. The one that has the potential to form the greatest number of ions will probably have the greatest number of disruption sites. You do not have the van’t Hoff constants, so the number of ions possible is the basis of discriminating among the salts. Note that they are all salts of group 1 metals and therefore highly soluble. It may also be assumed that they are strongly ionized, if not 100%, because all salts are “strong electrolytes”. (a) wrong. This only produces two ions, K+ and Cl . The molality of disruption sites = 0.4m, maximum. (b) wrong. Three ions, 2 Na + and one SO 4 2 . Molality of disruption sites = 0.6m, max. (c) correct . This mixture of salts has the potential to produce 2 ions from KCl and 3 ions from Na 2 SO 4 , giving molality of disruption sites = 1.0 m, max. (d) wrong. Sugar is non-ionizing and the molality of disruption sites = solute molality = 0.2m (e) wrong. The pure solvent will always have the HIGHEST vapor pressure possible. 3. Π V = nRT solve this relationship for n, the number of moles n = ) 292 )( / 363 . 62 ( ) 100 . 0 )( 1 . 28 ( K K mol L Hg mm L Hg mm RT V = Π = 1.5431 x 10 4 mol solute use mole and mass information to find molar mass of solute: molar mass = mol x g 4 10 5321 . 1 102 . 0 - = 663 g/mol ans. ( a) 4. By inspection of reaction stoichiometery, it is seen that 4 H 2 are produced every time one mole of C 6 H 14 is reacted. Therefore its production rate is 4 times faster than the disappearance rate of C 6 H 14 . To normalize rates to a “common rate” for the reaction, we multiply the disappearance of reactant species by ( 1). Rate of H 2 production = 4(rate of C 6 H 14 ) = 4 ( 6.2 x 10 3 ) = 2.48 x 10 2 M/s 2.5 x 10 2 M/s (2 sig figs) ans (c) 5. T bp = k bp m i = (3.33 o C/m)
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