CHM1046 Chapter 13: Solutions

Water is the solvent, and ethanol is the solute.

Greater solubility in hexane: toluene and isobutene. Greater solubility in water: sucrose and glycol. Polar solutes are more soluble in polar solvents (i.e., water), and nonpolar solutes are more soluble in nonpolar solvents (i.e., hexane). This is the basis of the rule of thumb that "like dissolves like."

dispersion. There are dispersion forces between toluene and hexane. These intermolecular interactions are the result of fluctuations in the electron distribution within molecules or atoms. Since the electrons in an atom or molecule may be unevenly distributed at any one instant, dispersion forces are present in all molecules and atoms. These forces increase in strength as molar mass increases because atoms or molecules with larger molar masses tend to have more electrons dispersed over a greater volume. No hydrogen-bonding, dipole-dipole, or ion-dipole intermolecular forces are present because neither toluene nor hexane contain any of the following: A hydrogen atom bonded to an electronegative element such as oxygen, nitrogen, or fluorine. Permanent dipoles. They do not exist in either hydrocarbon because only C−C, C=C, and C−H bonds exist, which do not have significant electronegativity differences between the atoms involved. Ionic bonds.

hydrogen bonding, dipole-dipole, and dispersion. There are dispersion forces between sucrose and water. These intermolecular interactions are the result of fluctuations in the electron distribution within molecules or atoms. Since the electrons in an atom or molecule may be unevenly distributed at any one instant, dispersion forces are present in all molecules and atoms. These forces increase in strength as molar mass increases because atoms or molecules with larger molar masses tend to have more electrons dispersed over a greater volume. There are also dipole-dipole intermolecular forces and hydrogen bonding taking place between sucrose and water because both molecules contain O−H bonds. Hydrogen bonding simply requires a hydrogen atom directly bonded to a small electronegative atom (i.e., oxygen), and there is a permanent dipole across the O−H bond because the atoms have significantly different electronegativities. Neither water nor sucrose are in the form of ions, so no ion-dipole interactions occur.

dispersion. There are dispersion forces between isobutene and hexane. These intermolecular interactions are the result of fluctuations in the electron distribution within molecules or atoms. Since the electrons in an atom or molecule may be unevenly distributed at any one instant, dispersion forces are present in all molecules and atoms. These forces increase in strength as molar mass increases because atoms or molecules with larger molar masses tend to have more electrons dispersed over a greater volume. No hydrogen-bonding, dipole-dipole, or ion-dipole intermolecular forces are present because neither isobutene nor hexane contain any of the following: A hydrogen atom bonded to an electronegative element such as oxygen, nitrogen, or fluorine. Permanent dipoles. They do not exist in either hydrocarbon because only C−C, C=C, and C−H bonds exist, which do not have significant electronegativity differences between the atoms involved. Ionic bonds.

hydrogen bonding, dipole-dipole, and dispersion. There are dispersion forces between ethylene glycol and water. These intermolecular interactions are the result of fluctuations in the electron distribution within molecules or atoms. Since the electrons in an atom or molecule may be unevenly distributed at any one instant, dispersion forces are present in all molecules and atoms. These forces increase in strength as molar mass increases because atoms or molecules with larger molar masses tend to have more electrons dispersed over a greater volume. There are also dipole-dipole intermolecular forces and hydrogen bonding taking place between ethylene glycol and water because both molecules contain O−H bonds. Hydrogen bonding simply requires a hydrogen atom directly bonded to a small electronegative atom (i.e., oxygen), and there is a permanent dipole across the O−H bond because the atoms have significantly different electronegativities. Neither ethylene glycol nor water are in the form of ions, so no ion-dipole interactions occur.

CH3OH ***CBr4 H2O NH3

Solvent-solvent interactions: interactions involving dipole-dipole attractions and interactions between the water molecules. Solute-solute interactions: interactions between the ions of sodium chloride and interactions involving ion-ion attractions. Solute-solvent interactions: interactions formed between the sodium ions and the oxygen atoms of water molecules and interactions formed between the chloride ions and the hydrogen atoms of water molecules. Correct! Explanation: The process of dissolution depends upon the separation of solute particles, the separation of solvent molecules, and the formation of the attractive interaction between the solute particles and the solvent molecules. The separation of solute particles (ΔH1) and the separation of solvent molecules (ΔH2) are endothermic process whereas the formation of the attractive interaction (ΔH3) is an exothermic process. Here, ΔH is the change in energy associated with each process. If the magnitude of ΔH3 is very large compared to ΔH1+ΔH2, then the process of dissolution occurs spontaneously and is exothermic. But if the magnitude of ΔH3 is very small compared to ΔH1+ΔH2, then the process of dissolution is endothermic and may occur depending upon the entropy change for the process. Specifically, the dissolution of NaCl is slightly endothermic. If you place several teaspoons of salt in a cup of water you can feel the glass getting cold.

***potassium fluoride, KF ***sodium iodide, NaI toluene, C7H8 ***lithium bromide, LiBr hexane, C6H14 Solution formation is favored by an increase in entropy. Unless solute-solute and solvent-solvent interaction are strong relative to solute-solvent interactions, a solution will form when an ionic substance is added to water. For a nonpolar substance, the solute-solvent interactions are weak relative to solute-solute and solvent-solvent interactions and, therefore, it will not form a solution in water.

Dipole-dipole forces: HF and CH3OH. Ion-dipole forces: AlCl3 and MgI2. Correct! Both CH3OH and HF have a hydrogen atom bonded to an electronegative atom. Such compounds can form hydrogen bonds with water molecules. Hydrogen bonds are a specific type of dipole-dipole attraction that is typically stronger.

Exothermic: forming solute-solvent attractions Endothermic: breaking solvent-solvent attractions and breaking solute-solute attractions

very exothermic slightly exothermic very endothermic ***slightly endothermic The dissolution of many salts, such as table salt, is slightly endothermic. The process of dissolving table salt, NaCl, in water has an enthalpy value of ΔHsoln=+4 kJ/mol. Therefore the temperature of the water will decrease slightly when the salt is added.

***|ΔHsolute| < |ΔHhydration| |ΔHsolute| > |ΔHhydration| |ΔHsolute| = |ΔHhydration| None of the above (nothing can be concluded about the relative magnitudes)

CH3COOH ***C6H5CH3 CH3OH C6H12O6

***CH3CH2CH2OH CH3CH2CH2CH2OH CH3CH2CH2CH2CH2CH2OH CH3CH2CH2CH2CH2OH

Some gas will bubble out of solution and more solid will dissolve.

Answers: -More gas molecules are soluble as pressure is increased. -The solubility of nitrogen gas at 2.00 atm is twice the solubility of the gas at 1.00 atm. -The solubility of a gas decreases with a decrease in pressure. -The concentration of gas particles in the solution is higher at 4.25 atm than at 1.00 atm. Explanation: Henry's law The solubility of a gas in a liquid solvent at a given temperature is directly proportional to the partial pressure of the gas over the solution. This relationship is expressed by a simple equation known as Henry's law: S=k×P where S is the solubility of the gas, P is the partial pressure of gas over the liquid, and k is the Henry's law constant. The Henry's law constant, k, is a value that depends on the specific gas and solvent at a specific temperature.

165 mg of N2 gas/100 g water

515 M Explanation: Using Henry's law equation, the solubility of ammonia gas is calculated as S===k×P58M/atm×8.88atm515M The solubility of gas in a liquid solvent depends upon the partial pressure of the gas over the solvent, temperature, and the specific type of solvent.

-Salt precipitates out of solution. ***-Gas bubbles out of solution. -Salt precipitates out of solution and gas bubbles out of solution. -Nothing happens; all of the salt and the gas remain dissolved in solution.

boiling point = 101.84 ∘C Explanation: The temperature change (ΔT) to the boiling point of a pure solvent depends on the molality (m) of the solute in the solution, and the equation is ΔT=m×Kb=3.6 m×0.512∘C/m Note that the boiling-point elevation constant of water (Kb=0.512∘C/m) differs from the freezing-point depression constant of water (Kf=1.86∘C/m), which is true for all solvents.

NaC2H3O2 KCl ***C6H12O6 Li2SO4

0.249 m ***0.338 m 0.216 m 0.0622 m

molarity = 0.600 M Molarity is the number of moles per liter of solution (mol/L). Therefore, calculating the number of moles in 1 L of solution allows you to determine the molarity, where 1 L of solution has a mass of 1030 g at the given density. The mass ratio (10.5%=0.105) multiplied by the solution mass (1030 g) yields the mass of solute (m=108g of C6H12O6). The mass of glucose is then converted to the number of moles using its molar mass (MW=180.156g/mol).

molality = 0.651 m Explanation: Molality is the number of moles per kilogram of solvent (mol/kg). Therefore, the mass ratio (10.5%=0.105) can be used to determine the mass of solute per unit mass of solution. Since a percent is the quantity per 100 units, you can deduce that 10.5 g of glucose is present in 100 g of solution. The mass of glucose is then converted to the number of moles using its molar mass (MW=180.156g/mol). The calculation of molality based on 1 kg of solution is m=n/mass solvent=(10.5 g C6H12O6)×(1 mol C6H12O6/180.156 g C6H12O6)(89.5 g H2O)×(1 kg H2O)/(1000 g H2O)

0.070 m LiBr 0.070 m C6H12O6 ***0.050 m Ca(NO3)2 either 0.070 m LiBr or 0.070 m C6H12O6

-mayonnaise only -shaving cream only ***-Both shaving cream and mayonnaise are colloids. -All three are colloids.