Chapter 11
A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous.
(a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K + and ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption.
(a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding
Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat.
Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals.
(a) Fe(NO 3 ) 3 is a strong electrolyte, thus it should completely dissociate into Fe 3 + and ions. Therefore, (z) best represents the solution. (b)
(a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved)
(a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces
The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating.
40%
2.8 g
2.9 atm
102 L HCl
The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region.
Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions.
(a) Find number of moles of HNO 3 and H 2 O in 100 g of the solution. Find the mole fractions for the components. (b) The mole fraction of HNO 3 is 0.378. The mole fraction of H 2 O is 0.622.
(a) (b) (c) (d)
In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.
(a) Determine the molar mass of HNO 3 . Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m
(a) 6.70 10 −1 m ; (b) 5.67 m ; (c) 2.8 m ; (d) 0.0358 m
1.08 m
(a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point. (b) 100.5 °C
(a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C
(a) Determine the molar mass of Ca(NO 3 ) 2 ; determine the number of moles of Ca(NO 3 ) 2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm
(a) Determine the molal concentration from the change in boiling point and K b ; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 10 2 g mol −1
No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by Δ T f = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is Δ T f = (1.0 m)(5.14 °C/ m ) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized.
144 g mol −1
0.870 °C
S 8
1.39 10 4 g mol −1
54 g
100.26 °C
(a) (b) Vapor pressures are: CH 3 OH: 55 torr; C 2 H 5 OH: 18 torr; (c) CH 3 OH: 0.75; C 2 H 5 OH: 0.25
The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C.
The observed change equals the theoretical change; therefore, no dissociation occurs.
| Colloidal System | Dispersed Phase | Dispersion Medium |
|---|---|---|
| starch dispersion | starch | water |
| smoke | solid particles | air |
| fog | water | air |
| pearl | water | calcium carbonate (CaCO 3 ) |
| whipped cream | air | cream |
| floating soap | air | soap |
| jelly | fruit juice | pectin gel |
| milk | butterfat | water |
| ruby | chromium(III) oxide (Cr 2 O 3 ) | aluminum oxide (Al 2 O 3 ) |
Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale.
If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate.