122pex1sol - Chemistry 122 - Su08 Solutions for Practice...

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Chemistry 122 - Su08 Solutions for Practice Midterm 1 1) Boyle’s Law P & V change at a constant T and n (moles) - V is indirectly proportional to P V % 1/P V = k(1/P) P*V = k (a constant) P 2 V 2 = P 1 V 1 or P f V f = P i V i 1 atm P 1 = 153 atm P 2 = 722 torr × ----------- = 0.950 atm 760 torr V 1 = 13.2 L V 2 = ? P 1 V 1 P 1 153 atm V 2 = -------- = (-----)V 1 = (--------------) 13.2 L = 2.13 x 10 3 L P 2 P 2 0.950 atm Check: If P dec, V inc and it did. A 2) Use Combined Gas Law : P f V f P i V i ------- = ------- T f T i P f = 2P i V f = 1/3 V i (Vol dec to 1/3 its initial value) P f V f P f V f T f = ------- T i = (-----)(-----) T i P i V i P i V i 2 P i 1/3 V i T f = (-------) (---------) T i = (2/3)T i P i V i C
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3) There are several methods for solving this problem. In all of them you need to know what STP is. STP => Standard Temp. & Pressure: T = 0°C (273.15K) & P = 1 atm a) Method 1 To get density (g/L) get moles from IGL (PV=nRT) first and convert to mass (grams). Then divide the mass by the volume. Assume 1 L (or just solve for n/V) PV (1 atm) (1 L) n = ------ = -------------------------------------------- = 4.46136 x 10 -2 mol RT (0.08206 L C atm/mol C K) (273.15 K) 32.00 g O 2 ? g = 4.46136 x 10 -2 mol × --------------- = 1.4276 g O 2 1 mol O 2 m 1.4277g O 2 D = ----- = ---------------- = 1.43 g/L V 1L b) Method 2 RT P = (------) D => D = (------) (in g/L) P RT (1 atm) ( 32.00 g/mol) D = ---------------------------------------------- = 1.42 g/L (0.08206 L C atm/mol C K) (273.15K) c) Method 3 (only usable at STP) Molar Volume at STP: V/n = 22.41 L/mol g 1 mol 32.00 g O 2 ? ---- = ----------- × --------------- = 1.42 g/L L 22.41 L 1 mol O 2 D
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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 n 2 if V 2 = V 1 then ---- = 1 ˆ n 2 = n 1 n 1 Substitute n = m/ = molar mass (molecular wt.) & m = mass m 2 m 1 2 m 2 ----- = ----- ˆ ----- = ----- The ratio of MW is same as ratio of masses 2 1 1 m 1 m 2 2 = (----- ) 1 m 1 m unknown 10.2 g 70.9 g unknown = (----------) Cl2 = (---------) (----------) = 28.0 g/mol (28.0 amu) m Cl2 25.8 g 1 mol B 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 = 1.00 atm; V = 120 mL = 0.120 L; T = 100°C + 273.15 = 37 3 .15 K PV (1.00 atm) (0.120 L) n CxHyOz = ------ = -------------------------------------------- = 3.9 1 85 x 10 -3 mol RT (0.08206 L C atm/mol C K) (37 3 .15 K) 0.345 g = ---------------------- = 88.04 g/mol or 88.04 amu 3.9 1 85 x 10 -3 mol EFW C2H4O = 44.05 amu 88.04 amu n = ---------------- = 1.9985 = 2 44.05 amu MF = (C 2 H 4 O) 2 = C 4 H 8 O 2 C
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6) 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 . P T = P O2 + P CO2 + P N2 a) Method 1 Use P i = P i P T & P i = n i /n Total n O2 0.115 mol P O2 = ------- = ------------------------------ = 0.096 8 0 n T (0.115 + 0.325 + 0.748) P O2 = P O2 P T = (0.096 8 0) (2.70 atm) = 0.26 1 36 atm = 0.261 atm a) Method 2 You could use the IGL along with the V, total P and total moles of gas to determine T. Then you could use the V, T and moles of CO 2 in the IGL to determine its partial pressure. This is the much harder way to do the problem. T = (PV)/(nR) = {(2.70 atm)(10.0 L)}/{(1.188 mol)(0.08206 L C atm/mol C K)} = 276.959 K P O2 = (nRT)/V = {(0.115 mol)(0.08206 L C atm/mol C K)(276.959 K)}/10.0L = 0.26 1 36 atm B 7) The density is directly proportional to the molar mass (molecular or atomic weight). Therefore, the gas with the smallest density should have the smallest molar mass (MW or AW) (since P, V & T same for all). The densities are listed below, as are the MW (which you didn’t need to calculate).
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This note was uploaded on 07/17/2008 for the course CHEM 122 taught by Professor Zellmer during the Summer '07 term at Ohio State.

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122pex1sol - Chemistry 122 - Su08 Solutions for Practice...

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