Chapter11rw-final

# Chapter11rw-final - 11 Intermolecular Forces Liquids and...

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298 11 Intermolecular Forces, Liquids, and Solids Visualizing Concepts 11.1 The diagram best describes a liquid . In the diagram, the particles are close together, mostly touching but there is no regular arrangement or order. This rules out a gaseous sample, where the particles are far apart, and a crystalline solid, which has a regular repeating structure in all three directions. 11.2 (a) (a) Hydrogen bonding; H–F interactions qualify for this narrowly defined interaction. (a) (b) London dispersion forces, the only intermolecular forces between nonpolar F 2 molecules. (a) (c) Ion-dipole forces between Na + cation and the negative end of a polar covalent water molecule. (a) (d) Dipole-dipole forces between oppositely charged portions of two polar covalent SO 2 molecules. (b) Ion-dipole forces in (a) (c) and hydrogen bonding in (a) (a) are stronger than the other two. 11.3 The viscosity of glycerol will be greater than that of 1-propanol. Viscosity is the resistance of a substance to flow. The stronger the intermolecular forces in a liquid, the greater its viscosity. Hydrogen bonding is the predominant force for both molecules. Glycerol has three times as many O–H groups and many more H-bonding interactions than 1-propanol, so it experiences stronger intermolecular forces and greater viscosity. (Both molecules have the same carbon-chain length, so dispersion forces are similar.) 11.4 (a) 385 mm Hg. Find 30 ° C on the horizontal axis, and follow a vertical line from this point to its intersection with the red vapor pressure curve. Follow a horizontal line from the intersection to the vertical axis and read the vapor pressure. (b) 22 ° C. Reverse the procedure outlined in part (a). Find 300 torr on the vertical axis, follow it to the curve and down to the value on the horizontal axis. (c) 47 ° C. The normal boiling point of a liquid is the temperature at which its vapor pressure is 1 atm, or 760 mm Hg. On this diagram, the vapor pressure curve ends at this point, approximately 47 ° C. 11.5 The stronger the intermolecular forces, the greater the average kinetic energy required to escape these forces, and the higher the boiling point. CH 3 CH 2 CH 2 OH has hydrogen bonding, by virtue of its –OH group, so it has the higher boiling point. Dispersion forces are similar because molar masses are the same for both molecules.

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11 Intermolecular Forces Solutions to Exercises 299 11.6 (a) 360 K, the normal boiling point; 260 K, normal freezing point. The left-most line is the freezing/melting curve, the right-most line is the condensation/boiling curve. The normal boiling and freezing points are the temperatures of boiling and freezing at 1 atm pressure. (b) The material is solid in the green zone, liquid in the blue zone, and gas in the tan zone. (i) gas (ii) solid (iii) liquid (c) The triple point, where all three phases are in equilibrium, is the point where the three lines on the phase diagram meet. For this substance, the triple point is approximately 185 K at 0.45 atm.
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