Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo

1.

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.

3.

(a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K+ and NO3NO3 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.

5.

(a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding

7.

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.

9.

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.

11.

(a) Fe(NO3)3 is a strong electrolyte, thus it should completely dissociate into Fe3+ and NO3NO3 ions. Therefore, (z) best represents the solution. (b) Fe(NO3)3(s)Fe3+(aq)+3NO3(aq)Fe(NO3)3(s)Fe3+(aq)+3NO3(aq)

13.

(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)

15.

(a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces

17.

The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating.

19.

40%

21.

2.8 g

23.

2.9 atm

25.

102 L HCl

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) XNa2CO3=0.0119;XNa2CO3=0.0119; XH2O=0.988;XH2O=0.988; (b) XNH4NO3=0.0928;XNH4NO3=0.0928; XH2O=0.907;XH2O=0.907; (c) XCl2=0.192;XCl2=0.192; XCH2CI2=0.808;XCH2CI2=0.808; (d) XC5H9N=0.00426;XC5H9N=0.00426; XCHCl3=0.997XCHCl3=0.997

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) XCH3OH=0.590;XCH3OH=0.590; XC2H5OH=0.410;XC2H5OH=0.410; (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.

Δbp=Kbm=(1.20°C/m)(9.41g×1mol HgCl2271.496g0.03275kg)=1.27°CΔbp=Kbm=(1.20°C/m)(9.41g×1mol HgCl2271.496g0.03275kg)=1.27°C
The observed change equals the theoretical change; therefore, no dissociation occurs.

71.
Colloidal System Dispersed Phase Dispersion Medium
starch dispersion starch water
smoke solid particles air
fog water air
pearl water calcium carbonate (CaCO3)
whipped cream air cream
floating soap air soap
jelly fruit juice pectin gel
milk butterfat water
ruby chromium(III) oxide (Cr2O3) aluminum oxide (Al2O3)
73.

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.

75.

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.

Order a print copy

As an Amazon Associate we earn from qualifying purchases.

Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction
Citation information

© Jan 8, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.