122practiceexamsolutions1

122practiceexamsolutions1 - Chemistry 122 W08 Solutions for...

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Chemistry 122 - W08 Solutions for Practice Midterm 1 1) P &V change at constant T - Use Boyle’s Law P 2 V 2 = P 1 V 1 or P f V f = P i V i P f = ? V f = 1/2 V i (Vol dec to 1/2 its initial value) P i V i V i P f = ------- = (-----) P i V f V f V i P f = --------- P i = 2 P i 1/2 V i C 2) Use Combined Gas Law : P, V, T all change D

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3) If you understand what standard molar volume is you can do this problem. The standard molar volume of a gas at our normal STP is 22.41 L/mol (1 mole of an ideal gas occupies 22.41 L). How was this value obtained? By simply plugging the P and T at STP into the ideal gas law and solving for V/n, the molar volume (volume per mole - same as solving for the volume of 1 mole of a gas). In this problem you are to define a new STP as defined by me in the problem: New STP => New Standard Temp. & Pressure: T = 20°C (293.15 K) & P = 700 mm Hg V RT (0.08206 L C atm/mol C K) (293.15 K) ---- = ------ = --------------------------------------------- n P 700 mm Hg x (1atm/760 mm Hg) = 26.117 L/mol = 26.1 L/mol You could also use the combined-Gas Law for 1 mole of a gas by using the standard molar volume of 22.41 L/mol at the normal STP (1 atm and 0°C) as your P 1 , V 1 and T 1 and solving for V 2 using the P and T given. E 4) According to Avogadro’s Law, equal volumes of gases under the same P & T conditions contain the same # moles (same # particles). V 2 V 1 n 2 ----- = ----- so V 2 = ---- V 1 n 2 n 1 n 1 V 1 = 503 mL V 2 = ? n 1 = 0.0243 mol n 2 = 0.0188 mol n 2 0.0188 mol V 2 = ---- V 1 = ---------------- (503 mL) n 1 0.0243 mol = 389.15 mL = 389 mL C
5) Determine molecular formula of a gas given the empirical formula, C 2 H 4 O, mass & P, V, T data MF = (EF) n &M W = n * (EFW) (where MW = molecular wt and EFW = empirical formula wt) MW mass n = --------- & = ---------- (molar mass or molecular wt) EFW moles Have mass of 0.345 g, need moles. (Remember, molar mass = g/mol, same as MW.) P = 0.942 atm; V = 85.0 mL = 0.0850 L; T = 100°C + 273.15 = 37 3 .15 K PV (0.942 atm) (0.0850 L) n CxHyOz = ------ = -------------------------------------------- = 2.6 1 48 x 10 -3 mol RT (0.08206 L C atm/mol C K) (37 3 .15 K) 0.345 g = ---------------------- = 13 1 .93 g/mol or 13 1 .93 amu 2.6 1 48 x 10 -3 mol EFW C2H4O = 44.05 amu 13 1 .93 amu n = ---------------- = 2.9951 = 3 44.05 amu MF = (C 2 H 4 O) 4 = C 6 H 12 O 3 D

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6) RT P = (------) D => D = (------) (in g/L) P RT The density is directly proportional to the molar mass (molecular or atomic weight). Therefore, the gas with the smallest molar mass (MW or AW) should have the lowest density (since P, V & T same for all). (a) (b) (c) (d) (e) Ne (20.18) CH 4 (16.04) Cl 2 (70.91) NH 3 (17.03) C 2 H 4 (28.05) B 7) Use Dalton’s Law of Partial Pressures, P Total = P 1 + P 2 + P 3 + CCC Each gas acts independently of the other gases and each follows the IGL as does P Tot . Open the valve and allow the gases to mix. Now each gas occupies the ENTIRE volume of both flasks, 4.0 L. Need to find new Pressure of each gas in “new” 4.0 L container - Use Boyle’s Law P He,1 V 1 (765 torr) (1.5 L) He P He,2 = ----------- = --------------------- = 28 6 .875 torr V 2 4.0 L P H2,1 V 1 (740 torr) (2.5 L) H 2 P H2,2 = ----------- = ---------------------- = 46 2 .5 torr V 2 4 .0 L P T = 28 6 .875 + 46 2 .5 = 74 9 .3 torr = 749 torr C
8) The rate of effusion is directly proportional to the molecular speed which is inversely proportional to the square root of the molar mass. Thus, lighter particles have a faster average velocity and faster rate of effusion. Have same # moles of H 2 & SO 2 in same container. H 2 effuses faster than SO 2 because it is lighter & thus has a faster average velocity and rate of effusion.

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This note was uploaded on 04/18/2008 for the course CHEM 121 taught by Professor Wyzlouzil during the Spring '07 term at Ohio State.

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122practiceexamsolutions1 - Chemistry 122 W08 Solutions for...

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