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Unformatted text preview: rahman (pr5989) – Homework 2 – flowers – (53880) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The normal boiling point of ethanol is 78 . 4 ◦ C. When 8 . 36 g of a soluble nonelectrolyte was dissolved in 86 g of ethanol, the vapor pres sure of the solution at that temperature was 711 Torr. What is the mole fraction of ethanol? Correct answer: 0 . 935526. Explanation: m solute = 8 . 36 g m ethanol = 86 g P = 711 Torr P pure solvent = 760 Torr At its boiling point, the vapor pressure of any liquid will be 760 Torr. P = X solvent × P pure solvent x solvent = P P pure solvent = 711 Torr 760 Torr = 0 . 935526 . 002 (part 2 of 2) 10.0 points What is the molar mass of the solute? Correct answer: 64 . 9793 g / mol. Explanation: FW C 2 H 5 OH = 2 (12 . 01 g / mol) + 6 (1 . 0079 g / mol) + 16 g / mol = 46 . 0674 g / mol . n solvent = 86 g 46 . 0674 g / mol = 1 . 86683 mol, so x solvent = n solvent n X + n solvent . 935526 = 1 . 86683 mol 8 . 36 g M X + 1 . 86683 mol 8 . 36 g M X = 1 . 86683 mol . 935526 1 . 86683 mol M X = 8 . 36 g 1 . 86683 mol . 935526 1 . 86683 mol = 64 . 9793 g / mol . 003 10.0 points What is the equilibrium expression for the following equation? A + 2 B ⇀ ↽ 3 C + D Assume that all species here are gases. 1. 3 [C] [D] 2 [A] [B] 2. [C] 3 [D] [A] [B] 2 correct 3. [A] [B] 2 [C] 3 [D] 4. 2 [A] [B] 3 [C] [D] Explanation: Equilibrium expressions are written with the products (raised to the powers of their co efficients) in the numerator and the reactants (raised to the powers of their coefficients) in the denominator. 004 10.0 points Calculate the vapor pressure at 25 ◦ C of a mixture of benzene and toluene in which the mole fraction of benzene is 0.650. The vapor pressure at 25 ◦ C of benzene is 94.6 torr and that of toluene is 29.1 torr. 1. 124 torr 2. 51.3 torr 3. 71.7 torr correct 4. 84.4 torr 5. 61.5 torr Explanation: 005 10.0 points rahman (pr5989) – Homework 2 – flowers – (53880) 2 What is the temperature change when 19 . 8 g of KNO 3 (with 34 . 9 kJ / mol enthalpy of solution) is dissolved in 50 g of water? As sume that the specific heat capacity of the solution is 4 . 18 J K · g . Correct answer: 32 . 7024 ◦ C. Explanation: m KNO 3 = 19 . 8 g m H 2 O = 50 g C = 4 . 18 J K · g MW KNO 3 = 101 . 103 g / mol Δ H = 34 . 9 kJ / mol = 34900 J / mol The number of moles is n = 19 . 8 g 101 . 103 g / mol = 0 . 195839 mol , so the heat released is obtained from n Δ H = mC Δ T Δ T = n Δ H mC Δ T = (0 . 195839 mol) (34900 J / mol) (4 . 18 J K · g ) (50 g) = 32 . 7024 K = 32 . 7024 ◦ C ....
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 Spring '09
 FLOWERS
 Chemistry, mol, Rahman

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