# 007 100 points version 111 exam 1 holcombe 52460 3

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Version 111 – Exam 1 – Holcombe – (52460) 3 What is the form of the equilibrium constant for the process PbO(s) + CO 2 (g) PbCO 3 (s) ? P species means the pressure of species. 1. K p = [PbCO 3 ] [PbO] P CO 2 2. No other choice is correct. 3. K p = P PbCO 3 P CO 2 4. K c = [PbCO 3 ] 2 [CO 2 ] [PbO] 5. K p = 1 P CO 2 correct Explanation: Solids are not included in the K expression. 008 10.0 points The observation that the solubility of gases in water decreases with increasing temper- ature means that the enthalpy of solvation H solvation )for gases is 1. zero 2. endothermic 3. greater than the change in entropy 4. exothermic correct Explanation: A high velocity gas molecule must release its excess kinetic energy as heat in order to decelerate and enter the aqueous phase. 009 10.0 points For a given transfer of energy, a greater change in disorder occurs when the temperature is high. 1. True 2. False correct Explanation: From Δ S = q T since T is in the denomina- tor, Δ S will be larger (more positive) when- ever T is smaller . 010 10.0 points Suppose the reaction A B has an equilibrium constant of 1.0 and the ini- tial concentrations of A and B are 0.5 M and 0.0 M, respectively. Which of the following is the correct value for the final concentration of A? 1. None of these is correct. 2. 0.500 M 3. 0.250 M correct 4. 1.00 M 5. 1.50 M Explanation: K = 1 . 0 [A] ini = 0 . 5 M [B] ini = 0 M A B ini, M 0.5 0.0 Δ, M - x x eq, M 0 . 5 - x x K = [B] [A] = 1 . 0 x 0 . 5 - x = 1 . 0 x = 0 . 25 M [A] = 0 . 5 - x = 0 . 25 M 011 10.0 points Which of the following two solutions will achieve the greatest boiling point increase and what will the new temperature be? Assume that K b = 0 . 52 C / m for water. I) 105 g sucrose (C 6 H 12 O 6 ) dissolved in 0.5 kg water II) 35 g of NaCl dissolved in 0.5 kg of water
Version 111 – Exam 1 – Holcombe – (52460) 4 1. Solution I with a boiling point of 101 . 92 C 2. Solution II with a boiling point of 101 . 25 C correct 3. Solution II with a boiling point of 100 . 63 C 4. Solution I with a boiling point of 100 . 32 C Explanation: First calculate the molality of each solution, which is mol solute per kg solvent. To do this, you will need to convert from mass to mols, using molar mass. Remember also that i = 2 for NaCl and i = 1 for sucrose. Using Δ T = imK b you will find that Δ T = 0 . 32 C for the sucrose solution and 1.25 for the salt solution, which means the boiling point of the salt solution will be 101.25 C. 012 10.0 points 30.2 g of glycerine (C 3 H 8 O 3 ) are dissolved in 150 g of water. What is the boiling point of the solution? ( K b of water = 0.515 C/m) 1. 100.10 C 2. 101.13 C correct 3. 0.104 C 4. 1.13 C 5. 103.52 C Explanation: m C 3 H 8 O 3 = 30.2 g m water = 150 g Δ T b = K b m = K b mol glycerol kg water = (0 . 515 C /m ) 30 . 2 92 . 1 mol C 3 H 8 O 3 0 . 150 kg water = 1 . 13 C T b = T 0 b + Δ T b = 101 . 13 C 013 10.0 points The vapor pressure of pure CH 2 Cl 2 (molecu- lar weight = 85 g/mol) is 133 torr at 0 C and the vapor pressure of pure CH 2 Br 2 (molecu- lar weight 174 g/mol) is 11 torr at the same temperature. What is the total vapor pres- sure of a solution prepared from equal masses of these two substances?

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