solution_pdf HW 2

solution_pdf HW 2 - rahman(pr5989 – Homework 2 –...

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Unformatted text preview: rahman (pr5989) – Homework 2 – flowers – (53880) 1 This print-out should have 30 questions. Multiple-choice 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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solution_pdf HW 2 - rahman(pr5989 – Homework 2 –...

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