122 HW_13 key

122 HW_13 key - g/w/WCL/ 13.3 Analyze/Plan. Decide whether...

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Unformatted text preview: g/w/WCL/ 13.3 Analyze/Plan. Decide whether the solute and solvent in question are ionic, polar covalent or nonpolar covalent. Draw Lewis structures as needed. Then, state the appropriate type of solute-solvent interaction. Solve: (a) CCI4, nonpolar; benzene, nonpolar; dispersion forces (b) CaClz, ionic; water, polar; ion-dipole forces (c) propanol, polar with hydrogen bonding; water, polar with hydrogen bonding; hydrogen bonding (d) HCI, polar; CHaCN, polar; dipole-dipole forces 13.6 Separation of solvent molecules, AHZ, will be smallest in this case, because hydrogen bonding is the weakest of the intermolecular forces involved. AH1 involves breaking ionic bonds, and AH3 involves formation of ion-dipole interactions, both stronger forces than hydrogen bonding. "N 10g KCIO3 13.12 a at 30°C, — X 250 H O = 25 KCIO ) 1OOgHZO 9 2 9 3 669Pb(N03)2 (b) —_ X 250 9 H20 = 165 = 1.7 X 102 g F’b(N03)2 1009H20 39C62(SO4)3 c ————-———X250 HO=7.5=8 Ce SO 1) 1OOgH20 9 2 9 2( 4):: 13.18 Hexane is a nonpolar hydrocarbon that experiences dispersion forces with other CsH14 ‘ molecules. Solutes that primarily experience dispersion forces will be more soluble in hexane. (a) Cyclohexane, CSHQ. is also a nonpolar hydrocarbon and will be more soluble in hexane. GlucOse experiences hydrogen bonding with itself; these solute-solute interactions are less likely to be overcome by weak solute-solvent interactions. (b) Propionic acid, which experiences hydrogen bonding, will be more soluble than sodium propionate, which experiences ion-ion forces. The hydrogen bonding is weaker (relatively speaking) and more likely to be overcome by dispersion forces with hexane. (c) Ethyl chloride, which has a —CHZCH3 group capable of dispersion forces, will be more soluble. HCI experiences dipole-dipole forces less likely to be disrupted by dispersion forces with hexane. Q 1 atm 13.22 665 torr x = 0.875 atm; P = p = 0.21 0.87 = I = . 760m 02 zo.( .) ( 5 atm) 01838 018 atm —3 so = kpo = W x 0.1333 atm =2_5 x104M 2 2 L-atm 13.24 (a) mass % = M x 100 J total mass solution 253.8 9 I2 =11.421 =11 I 1molI2 g 2 mass solute = 0.045 mol I2 x 11.421 9 I. 0 : ———————— mass “2 11.421912 + 11590Cl‘ X 100 = 9.034 = 9.0% [2 2* .z’V/V (b) ppm = mass solute x 106 = 0.0079 9 Sr x 10%: 7.9 ppm 8'2. ‘ total mass solution 1 x 103 9 H20 \ \\ 25.5 g C6H50H 94.1 1 g/mol 4959CH3CH20H 1 46‘07g/m0' — 01445-10] molCH3CH20H 13.26 (a) = 0.2710 = 0.271 mol CGHSOH 0.2710 = 0.02460 = 0.0246 25.5 g CGHSOH (b) mass % = = 255 g CGHSOH + 495 g CHacHZOH x 100 4.90% CSHSOH 0.2710mol C H OH (O) m = e 5 0.495 kg CHSCHZOH = 0.54747 = 0.547 m CSHSOH 13.40 Assume a solution volume of 1.00 L. Calculate the mass of 1.00 L of solution and the mass of HNO3 in 1.00 L of solution. 1.00 L x _1000 m x ____1‘42 950'“ = 1.42 x 103 g soln 1 L mL soln 16 mol HNO3 63.02 g HNO3 16 M: x =1008=1.0:<1039HNO3 1 L soln 1 mol HNO3 10089HNQ3 _ mass % = ————————— x 100 — 71% HNO3 1.42 x 103gsoln [13.4% v (a) H20 vapor pressure will be determined by the mole fraction of H20 in the solution. The (I ‘ ‘ .. vapor'pressure of pure H20 at 343 K (70°C) = 233.7 torr. 3 35.0 g C3H803 125 9 H20 ———— = 0.3800 = 0.380 mol; ~———— = 6.937 = 6.94 mol 92.10 g/mol 18.02 g/mol 6.937 mol H20 PH 0 - x 233.7 torr = 221.6 = 222 torr 2 _ 6.937 + 0.380 (b) Calculate 103 by vapor pressure lowering; x3 = APAIPA° (see Solution 13.47(b)). Given moles solvent, calculate moles solute from the definition of mole fraction. _ 10.0 torr -—~—— = 0.100 XCIH602 100 torr 1.00 x 103 g C2H5OH —————~——— = 21.71 = 21.7 mol CszoH; let y = mol CZHSOZ 46.07 g/mol y mol CZH602 = 0.100 — y y mol CZHSO2 + 21.71 mol CZHSOH ' y+21.71 XCZHGOZ — 0.100 y + 2.171 = y; 0.900 y = 2.171; y = 2.412 = 2.41 mol cszo2 2.412 mol (:szo2 x 6207 9 = 150 g czHeo2 1 mol 13.58 7T = MRT; T = 20°C + 273 = 293 K M(ofions) : mol NaCl X2 = 3.4gNaCI x 1molNaCl x 2mol Ions —— = 0.116 = 0.12 M L soln 1 L soln 58.4 9 NaCl 1 mol NaCl " = 0.116 mol X 0.08206 L-atm x 293 K = 2.8 atm L K-mol AT 13.60 AT,=5.5-4.1 = 1.4; m = 4 =i =027 = Kf 5‘12 . 3 0.27m 7% lauryl alcoho' = glauryl alcohol : 5.009lauryl alcohol \_ m x kg CGHS 0.273 x 0.100 kg CGHG = 1.8 x 102 g/mol lauryl alcohol rr : 6.41 atm x K-mol RT 298 K 0.08206 Loatm 13.87 M: = 0.2621 = 0.262 M .\‘/ There are 0.262 mol of particles per liter of solution and 0.131 mol particles in 500 mL. Each mole of sucrose provides 1 mol of particles and each mole of NaCl provides 2. Let x = g C12H220“, 15.0 - x = 9 NaCl , x g C12H22011 + 2(150 — x) W __—— 2 0.131 342.3 9 0121122011 58.44 g NaCl 58.44X + 2(342.3)(15.0 - x) = 0.131(58.44)(342.3) 58.44x — 684.6x + 10,269 = 2,621 —626.2x = -7648; x = 12.2 g C12H220“ 12.2 g C12H22011 x 100 = 81.3% sucrose, 18.7% NaCl 15.0 g mixture ...
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This note was uploaded on 07/17/2008 for the course CHEM 122 taught by Professor Zellmer during the Winter '07 term at Ohio State.

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122 HW_13 key - g/w/WCL/ 13.3 Analyze/Plan. Decide whether...

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