CHM 152 Chapter 11 Study Guide
Alloy
solid mixture of a metallic element and one or more additional elements
Exothermic Process
released when solute-solvent attractions are established (also known as solvation). Solution gets hotter
Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C. See Figure 11.17, and report the computed percentage to one significant digit.
(When looking at the figure, 10 degrees celcious lines up with 60 grams of KBr) mass % = grams of solute/total grams of solution x 100% = 60 g KBr / (60 g KBr + 100 g H20) x 100% = 37.5% rounded to three sig figs, the true answer is 40%
Which of the following gases is expected to be most soluble in water? Explain your reasoning. (a) CH4 (b) CCl4 (c) CHCl3
(a) CH4 will not be very soluable in water because the structure is very stable concluding that the gas in non-polar (b) CCl4 will not be very soluable in water because the structure is very stable concluding that the gas in non-polar (c) Is expected to be the most soluble because it is a polar gas. The molecular structure isn't stable.
Consider the solutions presented: (a) Which of the following sketches best represents the ions in a solution of Fe(NO3)3(aq)? In this figure, three beakers labeled x, y, and z are shown containing various arrangements of blue and red spheres suspended in solution. In beaker x, three small red spheres surround a single central blue sphere in small clusters which in turn are grouped in threes around a single red sphere, forming four larger clusters. In beaker y, the four large clusters are present without the central red spheres. Four individual red spheres are now present. In beaker z, the large clusters are not present. Twelve of the small clusters of three red and one blue sphere are present along with four single red spheres. (b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO3)3.
(a) Fe(NO3)3 is a strong electrolyte, thus it should completely dissociate into Fe3+ and (NO3−)(NO3−) ions. Therefore, (z) best represents the solution. (b) Fe(NO3)3(s)⟶Fe3+(aq)+3NO3−(aq)
What are the mole fractions of HNO3 and water in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
(a) Find number of moles of HNO3 and H2O in 100 g of the solution. Find the mole fractions for the components. (b) 68.0 g HNO3 100g - 68 = 32 g H2O The number of moles of HNO3 is 68 g/63.01 g/mol = 1.079 mol The number of moles of water is 32 g/18.015 g/mol= 1.776 mol. The mole fraction of HNO3 is 1.079/(1.079 +1.776) = 0.378. The mole fraction of H2O is 1 − 0.378 = 0.622
Compare the processes that occur when methanol (CH3OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.
(a) Methanol, CH3OH, dissolves in water in all proportions,interacting via hydrogen bonding.Methanol: CH3OH (l) + H20 --> CH3OH (aq) Hydrogen chloride, HCl, dissolves in and reacts with water to yield hydronium cations and chloride anions that are solvated by strong ion-dipole interactions.Hydrogen chloride: HCL (g) + H20 --> H3O+(aq) + Cl- (aq) Sodium hydroxide, NaOH, dissolves in water and dissociates to yield sodium cations and hydroxide anions that are strongly solvated by ion-dipole interactions and hydrogen bonding, respectively.Sodium hydroxide: NaOH(s)--> Na+(aq) + OH-(aq)
When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally. (a) Is the dissolution of KNO3 an endothermic or an exothermic process? (b) What conclusions can you draw about the intermolecular attractions involved in the process? (c) Is the resulting solution an ideal solution?
(a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K+ and NO3−NO3− 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.
Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C7H16, nonpolar solvent): (a) vegetable oil (nonpolar) (b) isopropyl alcohol (polar) (c) potassium bromide (ionic)
(a) heptane (b) water (c) water
What is the expected electrical conductivity of the following solutions? (a) NaOH(aq) (b) HCl(aq) (c) C6H12O6(aq) (glucose) (d) NH3(l)
(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)
Indicate the most important types of intermolecular attractions in each of the following solutions: (a) The solution in Figure 111.2 (b) NO(l) in CO(l) (c) Cl2(g) in Br2(l) (d) HCl(aq) in benzene C6H6(l) (e) Methanol CH3OH(l) in H2O(l)
(a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding
Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions: (a) the solutions in Figure (b) methanol, CH3OH, dissolved in ethanol, C2H5OH (c) methane, CH4, dissolved in benzene, C6H6 (d) the polar halocarbon CF2Cl2 dissolved in the polar halocarbon CF2ClCFCl2 (e) O2(l) in N2(l)
(a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces
How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? What is the freezing point of this solution?
3.08 m = 3.08 mole glycerin/1 kg water First, find the molar mass of glycerin. The molar mass of glycerin is 92.095 g. A 3.08m aqueous solution requires 3.08 mol * (92.095 g/1 mole) = 284g glycerin To prepare the solution, dissolve 284 g of glycerin in 1.00 kg of water. Glycerin is nonelectrolyte. ΔTf = 1.86 °C/m * 3.08 m = 5.73 (1.86 is the freezing point depression constant for glycerin) The freezing point is -5.73 °C.
Calculate the mole fractions of methanol, CH3OH; ethanol, C2H5OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.)
Assume that the total solution mass is 100 g. Calculate the mass and the number of moles of each component. Then calculate the respective mole fractions. mol CH3OH = 0.40 * 100 g * (1mol/32.042 g) = 1.24836 moles mol C2H5OH = 0.40 * 100 g * (1 mol/46.069 g) = 0.86827 moles mol H2O = 0.20 * 100 g * (1 mol/18.0152 g) = 1.1101736 mol Total number of moles = 1.24836 mol + 0.84827 mol + 1.1101736 mol = 3.22680 mol Fractions: X CH3OH = 1.24836/3.22680 = 0.39 X C2H5OH = 0.86827/3.22680 = 0.27 X H2O = 1.1101736/3.22680 = 0.34
Explain why the ions Na+ and Cl− are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.
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.
What is the boiling point of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water?
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. mol surose = 115.0 g/342.300 g = 0.3360 mol mol molality = 0.3360 mol C6H12O6/0.3500 kg H2Om = 0.9599 m
Cg=kPg
Henry's Law equation The relation between solubility, Cg, and partial pressure, Pg, is a proportional one: where k is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature.
Footnote 1
If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution
Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3. How could you determine whether the solution is unsaturated, saturated, or supersaturated?
If you add a little bit of sodium thiosulfate to an unstaurated solution, the crystals would dissovle. If you add a little bit of sodium thiosulfate to a saturated solution, the added crystals would sink to the bottom. If you add a little bit of sodium thiosulfate to a supersaturated solution, the added crystals would cause a lot of the other crystals to fall and make the solution cloudy.
What is the difference between a 1 M solution and a 1 m solution?
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.
What is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?
Molality = moles of solute/kg of solvent 125 g H2O = 1 kg/ 1000g = 0.125 kg H2O 14.5 g H3PO4 = 1 mole/97.9952 g H3PO4 = 0.1479 moles H3PO4 m = 0.1479 mole/0.125 kg = 1.18 m
The concentration of glucose, C6H12O6, in normal spinal fluid is 75mg/100g. What is the molality of the solution?
Molality = moles of solute/kg of solvent 75 mg * 1 g/ 1000 mg = 0.075 grams C6H12O6 Molar Mass of C6H12O6 = 180.16 grams 0.075 grams * 1 mole/180.16 grams C6H12O6 = 0.0004163 moles C6H12O6 100 g * 1 kg/1000 grams = 0.1 kg molality = 0.0004163 moles/0.1 kg = 0.0042 m
What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
Molality = moles of solute/kg of solvent Molar mass HNO3= 63.01288 g If we assume 100 g of solution, then 68.0 g is HNO3 and 32.0 g is water. mol HNO3 = 68.0 g HNO3 * (1 mol/63.02188g HNO3 = 1.08 mol molality HNO3 = 1.08 mole/0.0320 kg = 33.7 m
Solutions vs Mixtures
Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous
Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and that ethanethiol, C2H5SH, is soluble only to the extent of 1.5 g per 100 mL of water.
The hydrogen bonds between water and ethanol are much stronger than the intermolecular bonds between water and ehanethiol. Because of the strong hydrogen bond attractions between ethanol and water, large amounts of ethanol can still be completely miscible with the water. Whereas, only a small amount of ethanethioil can be homoegously mixed with water.
Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin?
The presence of two ions for each NaCl molecule in a solution of NaCl compared with only one molecule for each glycerin in a solution of glycerin doubles the freezing point depression. This relationship is accounted for by the van't Hoff factor in the expression ΔTf= i Kfm where "i" is the van't Hoff factor and counts the number of distinguishable particles after the solute dissolves in the sol
A solution contains 5.00 g of urea, CO(NH2)2, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution?
The vapor pressure of the pure solvent is known; therefore, Raoult's law can be used along with the mole fraction to calculate the vapor pressure of the solution. First, calculate the molar amounts of urea and water, then compute the mole fraction of water, then use Raoult's law to compute the solution's vapor pressure: Psolution = Xsolvent * Psolvent X H20 = 5.55/ (0.833 + 5.55)= 0.9852 P solution = (0.9852)(23.7 torr) = 23.3 torr Remember that the mole fraction in this equation is related to the mole fraction of the solvent rather than the mole fraction of the solute
Solvation
exothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established
solubility
extent to which a solute may be dissolved in water, or any solvent
unsaturated
of concentration less than solubility
supersaturated
of concentration that exceeds solubility; a nonequilibrium state
partially miscible
of moderate mutual solubility; typically refers to liquid substances
immiscible
of negligible mutual solubility; typically refers to liquid substances
Spontaneous Process
physical or chemical change that occurs without the addition of energy from an external source
Dissociation
physical process accompanying the dissolution of an ionic compound in which the compound's constituent ions are solvated and dispersed throughout the solution
Ideal Solutions
solution that forms with no accompanying energy change
Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na2SO4, and 0.030 mol of MgCl2, assuming complete dissociation of these electrolytes.
ΔT = (K)(total molaity of all particles) ΔT = k*m 0.010 mol NaCl contains 0.010 mol Na + 0.010 mol Cl = 0.020 mol 0.020 mol Na2SO4 contains 0.040 mol Na + 0.020 mol SO4 = 0.060 mol 0.030 mol MgCl2 contains 0.030 mol Mg2 + 0.060 mol Cl = 0.090 mol Total numbers of moles = 0.020 mol + 0.060 mol + 0.090 mol = 0.170 mol molality = 0.170 moles /.100kg of water = 0.170 ΔT= 0.512 * 0.170 = 0.870° C (0.512 ° C/m is the boiling point constant)
Solutions vs Compounds
A solution can vary in composition, while a compound cannot vary in composition
What are the mole fractions of H3PO4 and water in a solution of 14.5 g of H3PO4 in 125 g of water?
14.5 grams H3PO4 = 1 mole/97.9952 grams H3PO4 = 0.148 mole H3PO4 125 grams H2O = 1 mole/ 18.0153 grams H2O = 6.939 mole H2O Total number of moles: 0.148 H3PO4 + 6.939 H2O = 7.087 moles Fraction: X H3PO4 = 0.148 mole H3PO4/7.087 = 0.0209 X H2O= 6.939 mole H2O/7.087 = 0.979
A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C3H5(OH)3), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical?
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.
Arrange the following solutions in order by their decreasing freezing points: 0.1 m Na3PO4, 0.1 m C2H5OH, 0.01 m CO2, 0.15 m NaCl, and 0.2 m CaCl2.
Order from least molality to greatest molality: 1) CO2 ---> CO2 (aq) 1 ion 1 ion --> 0.01m x 1 = 0.01 m 2) C2H5OH ---> 1 C2H5OH(aq) 1 ion --> 0.1m x 1 = 0.1 m 3) NaCl ---> Na + 2 Cl 3 ions --> 0.15m * 3 = 0.3m 4) Na3PO4 ---> 3 Na + 1 PO4 4 ions --> 0.1 m x 4 = 0.4m 6) CaCl2 ---> Ca + 2Cl 3 ions --> 0.2m x 3 = 0.6 m
What is the microscopic explanation for the macroscopic behavior of a glass containing oil and water?
The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Water is polar and attracts polar molecules, oil is non-polar and attracts non-polar 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 an interface.
ion-dipole attraction
electrostatic attraction between an ion and a polar molecule
Henry's Law
law stating the proportional relationship between the concentration of dissolved gas in a solution and the partial pressure of the gas in contact with the solution
Solutions
may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces
miscible
mutually soluble in all proportions; typically refers to liquid substances
saturated
of concentration equal to solubility; containing the maximum concentration of solute possible for a given temperature and pressure
strong electrolyte
substance that dissociates or ionizes completely when dissolved in water
nonelectrolyte
substance that does not produce ions when dissolved in water
weak electrolyte
substance that ionizes only partially when dissolved in water
Electrolyte
substance that produces ions when dissolved in water
Endothermic Process
the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions. Solution gets colder
Solvent
the most concentrated component and determines the physical state of the solution
Formation of a solution
when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level