| 11.4 Colligative Properties |
| 27. |
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. |
| 29. |
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. |
| 31. |
(a) |
Find number of moles of HNO3 and H2O in 100 g of the solution. Find the mole fractions for the components. |
|
(b) |
The mole fraction of HNO3 is 0.378. The mole fraction of H2O is 0.622. |
| 33. |
(a) |
[latex]X_{\text{Na}_2\text{CO}_3}= 0.0119[/latex] |
|
|
[latex]X_{\text{H}_2\text{O}}= 0.988[/latex] |
|
(b) |
[latex]X_{\text{NH}_4\text{NO}_3}= 0.0928[/latex] |
|
|
[latex]X_{\text{H}_2\text{O}}= 0.907[/latex] |
|
(c) |
[latex]X_{\text{Cl}_2}= 0.192[/latex] |
|
|
[latex]X_{\text{CH}_2\text{Cl}_2}= 0.808[/latex] |
|
(d) |
[latex]X_{\text{C}_5\text{H}_9\text{N}}= 0.00426[/latex] |
|
|
[latex]X_{\text{CHCl}_3}= 0.997[/latex] |
| 35. |
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. |
| 37. |
(a) |
Determine the molar mass of HNO3. 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 |
| 39. |
(a) |
6.70 × 10−1 m |
|
(b) |
5.67 m |
|
(c) |
2.8 m |
|
(d) |
0.0358 m |
| 41. |
1.08 m |
| 43. |
(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 |
| 45. |
(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 |
| 47. |
(a) |
Determine the molar mass of Ca(NO3)2; determine the number of moles of Ca(NO3)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 |
| 49. |
(a) |
Determine the molal concentration from the change in boiling point and Kb; 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 × 102 g mol−1 |
| 51. |
No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by ΔTf = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is ΔTf = (1.0 m)(5.14 °C/m) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized. |
| 53. |
144 g mol−1 |
| 55. |
0.870 °C |
| 57. |
S8 |
| 59. |
1.39 × 104 g mol−1 |
| 61. |
54 g |
| 63. |
100.26 °C |
| 65. |
(a) |
[latex]X_{\text{CH}_3\text{OH}}= 0.590[/latex] |
|
|
[latex]X_{\text{C}_2\text{H}_5\text{OH}}= 0.410[/latex] |
|
(b) |
Vapor pressures are: CH3OH: 55 torr; C2H5OH: 18 torr |
|
(c) |
CH3OH: 0.75; C2H5OH: 0.25 |
| 67. |
The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C. |
| 69. |
[latex]\Delta\text {bp} = K_bm = (1.20^{\circ}C/m)(\frac{\text {9.41 g}\times{\frac{\text {1 mol HgCl}_2}{\text {271.496 g}}}}{\text {0.03275 kg}}) = 1.27^{\circ}C[/latex] |
|
The observed change equals the theoretical change; therefore, no dissociation occurs. |
