INTERMOLECULAR FORCES AND COLLIGATIVE PROPERTIES HOMEWORK ANSWERS

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Oswald Piers Jacobs
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1 INTERMOLECULAR FORCES AND COLLIGATIVE PROPERTIES OMEWORK ANSWERS a)-c) intermolecular d) intramolecular a), b) intermolecular; c), d) intramolecular No. At 1.1 atm, water boils at a temperature above 100 o C, since it is more difficult for gas molecules to escape the liquid when the applied pressure is greater ln(p 2 /P 1 ) = VAP 1 1 R T2 T1 ln(760/p 1 ) = 1.5 x 10 J/mol J/mol K 9 K 1 8 K ln(760/p 1 ) = /P 1 = 1.44 P 1 = 528 torr ln(p 2 /P 1 ) = VAP 1 1 R T2 T1 R(ln( P2 / P1)) (8.14 J/mol K) (ln(760 / 640)) VAP = = K 58 K T2 T2 = (8.14) (0.170) J/mol (-7.9 x 10 5 ) = 1.8 x 10 4 J/mol (= 18 kj/mol) P, atm T, K ydrogen does sublime at 0.05 atm, since 0.05 atm is below the triple-point pressure.

2 Intermolecular forces involve interactions of lower (partial) charges at relatively larger distances than in covalent bonds Even though molecules are neutral, many of them are polar. These polar molecules will orient themselves so that their partial charges will result in dipole-dipole interactions. The partially positive pole of one molecule attracts the partially negative pole of another All molecules exhibit dispersion forces. Since these are so weak, however, they are in general overwhelmed if any other intermolecular forces (dipole-dipole, -bonding) exist If the electron distribution in one molecule is unsymmetrical (permanent or temporary), that can induce a temporary dipole in an adjacent molecule by causing the electrons in that molecule to shift for some (perhaps short) time a)-bonding b) dispersion c) dispersion a) -bonding b) dipole-dole c) ionic bonds a) dipole-dipole b) dispersion c) -bonding a) dispersion b) dipole-dipole c) -bonding C.. N C O C C.. O O C C.. O C.. N C a) dispersion b) -bonding c) dispersion a) C 2 6. Smaller molecules have smaller intermolecular forces. b) C C 2 F. Weaker intermolecular force (dipole-dipole vs. -bonding in C C 2 O). c) P. Weaker intermolecular force (dipole-dipole vs. -bonding in N ) a) OC 2 C 2 O. Larger intermolecular force (more O groups to -bond). b) C COO. Larger intermolecular force (-bonding vs. dipole-dipole). c) F. Larger intermolecular force (-bonding vs. dipole-dipole) a) LiCl; ionic bonds (LiCl) vs. dipole-dipole (Cl). b) N ; -bonding (N ) vs. dipole-dipole (P ). c) I 2 ; larger molecule has larger dispersion forces a) C C 2 O: -bonding (C C 2 O) vs. dispersion (C C 2 C ). b) NO; dipole-dipole (NO) vs. dispersion (N 2 ). c) 2 Te; larger molecule has larger dispersion forces C C 2 C 2 O < OC 2 C 2 O < OC 2 C(O)C 2 O

3 The number of -bonds, and hence the strength of intermolecular forces, would increase as shown, so also would the surface tension Viscosity and surface tension both increase with the strength of intermolecular forces, so the order would be the same Water is a good solvent for materials which form ions and/or which can hydrogen bond. The more polar (i.e., like water) the material, the better. Water is a poor solvent for nonpolar materials The atom in 2 O has four pairs of electrons around it. In 2 O itself, two are covalent bonds and two are lone pairs. The two lone pairs can form hydrogen bonds with suitable other molecules, making a total of four "bonds." a) ionic bonds between Na + and Cl and -bonds between water molecules. b) dispersion forces between Kr atoms. c) -bonds between water molecules. d) dispersion forces between CO 2 molecules a) A: solid E: solid + liquid F: liquid + gas : liquid B: liquid + solid + gas C: gas b) critical point: D triple point: B c) BG d) The substance is a solid, which would melt and then boil. e) The substance is a liquid which would vaporize a) In the polar solvent water, KNO ionizes completely and forms more concentrated solutions. KNO, an ionic salt, does not dissolve in or interact well with the non-polar carbon tetrachloride a) hydrogen bonding b) dipole-induced dipole c) dispersion forces a) dispersion forces b) hydrogen bonding c) dispersion forces 10.0 g CO x 1 mol CO / 2.04 g C O a) C(m)C O = =.12 m kg 2O m particles =.12 m 20.0 g CC2O x 1 mol CC2O / g CC2O C(m)C C 2 O = kg 2O m particles = 2.17 m 10.0 g of C O in 100. g of 2 O has the lower freezing point. b) C(m) 2 O = m particles = m 20.0 g 2O x 1 mol 2O /18.02 g 2 O = m 1.00 kg C O = 2.17 m

4 10.0 g CC2O x 1 mol CC2O / g CC2O C(m)C C 2 O = 1.00 kg CO m = m particles = m 10.0 g 2 O in 1.00 kg C O has the lower freezing point a) C(m)C 8 O = m particles = 1.52 m C(m)C 2 6 O 2 = m particles = 2.26 m 5.0 g CO 8 x ( 1mol CO 8 /92.09 g CO 8 ) kg ethanol 5.0 g C26O2 x ( 1 mol C26O 2 / g C26O2) kg ethanol = 2.26 m = 1.52 m The solution of 5.0 g of C 2 6 O 2 in 250 g of ethanol would have the higher boiling point. b) C(m)C 2 6 O 2 = 20. g CO x 1 mol CO /62.07 g CO 0.50 kg O = 0.64 m m particles = 0.64 m C(m)NaCl = 20. g NaCl x 1 mol NaCl / g NaCl 0.50 kg O 2 = 0.68 m m particles = 1.4 m (2 mol of ions per mol NaCl) 20. g of NaCl in 0.50 g of 2 O has the higher boiling point I m NaNO = m ions II m glucose = m molecules III m CaCl 2 = 0.00 m ions a) osmotic pressure: I and II same; III higher b) boiling point: I and II increased; III increased by 1½ times c) freezing point: I and II decreased; III decreased by 1½ times d) vapor pressure: I and II decreased; III decreased by 1½ times I 0.04 m ( 2 N) 2 CO II 0.02 m AgNO All have 0.04 m particle concentrations and have the III 0.02 m CuSO 4 same colligative properties g C8O g / mol = mol C 8 O g 2O = mol 2O g / mol P solvent = X solvent x P solvent P solvent = mol ( ) mol 2.76 torr P solvent = x 2.76 = 2.6 torr

5 T f = K f m = (1.86 C/m)(0.111 m) T f = C, so the f.p. of m urea solution is C T b = K b m = (0.52 C/m)(0.200 m) T b = C, so the b.p. of m lactose is C n(mol) vanillin =.4 g vanillin g / mol vanillin = mol vanillin (solute) 50.0 g ethanol m(kg) ethanol = 10 g / kg C(m) = mol kg = 0.45 m T b = K b m = (1.22 C/m)(0.44 m) = 0.54 C b.p. = 78.5 C C = 79.0 C = kg ethanol (solvent) n(mol) = 5.00 g C 10 8 x 1 mol g = mol C kg m(kg) = 444 g benzene x = kg benzene 10 g mol C(m) = = m kg T f = K f m = (4.90 C/m)( m) T f = 0.40 f.p. = 5.5 C 0.40 = 5.1 C C = 5 9 (F 2) = 5 9 ( ) = 2. C mol of solute T f = K f m = K f kg of solvent Tf kg mol of solute = ik f n (mol) C 2 6 O 2 = ( 2. C)( 14.5 kg) ()( C/m) = 182 mol m (kg) C 2 6 O 2 = 182 mol x g 1 mol 1 kg x = 11. kg C2 6 O 2 10 g

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