solute
Chapter 13
Properties of Solutions
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The Solution Process A solution is a ho m o g e n e o u s m i x t u r e of solute (present in smallest amount) and solvent (present in largest amount).
→ Solutes and solvents are THE components of the solution.
→ A SATURATED SOLUTION contains the maximum amount of solute.
→ Solubility depends on intermolecular forces. In the process of making solutions, intermolecular forces become rearranged.
Consider NaCl (solute) dissolving in water (solvent):
→ The water H-bonds have to be interrupted, → NaCl dissociates into Na+(aq) and Cl-(aq), → io n -di p o l e forces form, → We say the ions are
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-3--
Factors Affecting Solubility Solute-Solvent Interactions
Polar liquids tend to dissolve in polar solvents, (LIKE DISSOLVES LIKE).
Miscible liquids: MIX in any proportions. Immiscible liquids: DO NOT MIX. Intermolecular forces are important: water and ethanol are miscible because hydrogen bonding in both pure liquids are re-established in the mixture.
Fa c t o r s to consider in deter mining solubility in various SOLVENTS
1. The NUMBER of carbon atoms in a chain affect solubility: the m o r e C atoms, the LESS soluble in water.
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2. The number of -OH groups within a molecule inc r e a s e s solubility in water (via hydrogen bonding).
Structure of glucose. Note that OH groupS capable of hydrogen bonding with water are prominent on the "surface" of the molecule.
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3. The m o r e po l a r bo n d s in the molecule, the better it dissolves in a polar solvent. 4. The le s s po l a r t h e m o l e c u l e the less it dissolves in a polar solvent and the
better is dissolves in a non-polar solvent.
→
Generalization: "like dissolves like".
EX: Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride, CCl4, which is nonpolar, or in water, which is polar: C7H16, NaHCO3, HCl, I2. SOLUTION:
→ →
Both C7H16 and I2 are nonpolar. We would therefore predict that they would be more soluble in CCl4 than in H2O. NaHCO3 is ionic, and HCl is polar covalent. Water would be a better solvent than CCl4 for these two substances.
-6--
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Naphthalene, sometimes used as moth crystals, has the chemical formula C10H8. It is a ring structure that looks a lot like two hexagonal benzene rings sharing a common two-carbon side. In which of the following solvents will naphthalene dissolve the best?
a) H2O(l) b) CH3OH(l) (methanol) c) C6H6(l) (benzene) d) NH3(l) e) CH3CH2OH(l) (ethanol)
-7--
Factors Affecting Solubility Solubility of gas depends on Pressure Effects
Solubility of a gas in a liquid increases in direct proportion to its partial pressure (relative to the gases in the atmosphere or surroundings) above the solution.
The hi g h e r t h e pr e s s u r e s, the more molecules of gas are clo s e to the solvent and the greater the chance of a gas molecule striking the surface and entering the solution. Therefore, the higher the pressure, the greater the solubility.
The lower the pressure, the fewer m olecules of gas are close to the solvent, and the lower the solubility.
-
-8--
Henry’s Law The relationship between PRESSURE and the SOLUBILITY of a GAS is expressed by the following simple equation:
Sg = kPg Sg is the solubility of the gas (mol/L) k is Henry’s law constant (mol/L-atm), which is unique for each solute-solvent pair
Pg is the partial pressure of the gas above the solvent
Example: Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over liquid at 25°C. The Henry’s Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm. Sg = kPg = (3.1 x 10-2 mol/L-atm)( 4.0 atm) Sg = 0.12 mol/L or 0.12 M
Temperature Effects on Solubility
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-9--
→ → → → →
Experience tells us that sugar dissolves better in warm water than cold. As t e m p e r a t u r e inc r e a s e s , solubility of solids generally increases. Sometimes (rarely), solubility decreases as tem perature increases (e.g., Ce 2 (SO 4 )3 ). Experience tells us carbonated beverages go flat as they get warm.
Thermal pollution: if lakes get too warm, CO2 and O2 become les s sol u b l e and are not available for plants or animals.
Mole Fraction, Molarity, and Molality
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- 10 - -
We often use concentration expressions based on the number of moles of one or more components of the solution.
The three we use most commonly are mole fraction, Molarity, and molality.
We introduced Molarity (M) in chapter 4. Recall that the Molarity of a solute in a solution is defined as:
Recall from chapter 10, that the mole fraction of a component of a solution is given by:
or
→
Xsolute = nsolute/ nsolute + nsolvent
The symbol, X, is commonly used for mole fraction, with a subscript to indicate the component of interest.
→ NEW! The MOLALITY of a solution, denoted m, is defined as the number of moles of solute per kil o g r a m (DON’T FORGET) of SOLVENT:
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- 11 - -
EX: A solution of hydrochloric acid contains 36 grams of HCl and 64 grams of water. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution. SOLUTION (a) The solution contains 36 g of HCl and 64 g of H2O. To calculate the mole fraction of HCl, XHCl, we first convert mass to moles: mol’s of HCl = 36g/36.5 gmol-1 = 0.986 mol mol’s of H2O = 64g/18 gmol-1 = 3.56 mol XHCl = 0.986/0.986 + 3.56 = 0.22 (b) To calculate the molality of HCl in the solution, we use water is the solvent. We calculated the number of moles of HCl in part (a), and the mass of solvent is 64 g = 0.064 kg. Don’t forget to convert!! m of HCl = 0.986 mol/0.064 kg = 15.4 m
Colligative Properties What is it?
FREEZING POINT LOWERING and BOILING POINT ELEVATION are physical properties of solutions that depend on the quantity (molarity or molality) of molecules, but not the kind or identity of solute particles.
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- 12 - -
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One consequence of these colligate properties results from: LOWERING VAPOR PRESSURE
Vapor pressure in liquids goes do w n w he n nonvolatile (don’t boil off easily)solutes are dissolved in the m .
Nonvolatile contaminants (ie. Solute) make liquids harder to boil and freeze [EX: antifreeze/water mixtures],
The extent to which a nonvolatile solute lowers its vapor pressure (harder to boil - less vaporneed more heat) is proportional to its concentration. The relationship expressed by Raoult’s Law.
Raoult’s Law := states that the partial
pressure exerted by a solvent vapor above a solution, PA, equals the product of the mole fraction of the solvent in the solution, XA, times the vapor pressure of the pure solvent, P°A.
P A = X A P °A
Example: The vapor pressure of water is 17.5 torr at 20°C. At constant temperature you add
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sugar to water so that the resulting soltution has a 0.8 mol of water and 0.4 mol of sugar. Then the vapor pressure of water the solvent should be lowered making it boil at a higher temperature. Calculate: We are interested in the vapor pressure of the solvent, water, need XH2O: XH2O = nH2O/ nsugar+ nwater = 0.8/0.4 + 0.8 = 0.67 PA = XA P°A = PH2O = 0.67(17.5 torr) = 11.725 torr The presence of the nonvolatile substance (sugar) lowered the vapor pressure of the solvent 5.8 torr! Requiring more heat to boil the solution!
Also, bear in mind from chapter 11, that intermolecular forces are at play! (H-bonding in the above example)
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- 14 - -
-
Te m p e r a t u r e shi f t ,∆T, of mp or bp from a NORMAL value is proportional to m ole fraction of the conta minant dissolved in an nonvolatile solution. (boiling point is increased or freezing point lowered)
decreased
increased
The INCREASE in boi l i n g poi n t relative to that of the pure solvent, ∆Tb, is directly proportional to the number of solute particles per mole of solvent molecules.
∆Tb is proportional to molality: ∆Tb = iKb m i = van't Hoff factor For molecular solutes, i = 1 For ionic solutes, i = total number of cations and anions (ex., i = 3 for MgCl2 )
- 15 - -
Kb := is the molal boiling point elevation constant
-
(see table below)
m := molality expresses the number of moles of solute per 1000 g of solvent.
Like the boiling-point elevation, the DECREASE in freezing point, ∆Tf, is directly proportional to the molality of the solute: ∆ Tf = iKf m
New Boiling pt of soln. = normal bp + ∆ Tb (increased) New freezing pt of soln. = normal fp - ∆ Tf (decreased)
Sugar Example (Molecular, I =1): For water, Kb is 0.52°C/m; therefore, a 1.0 m aqueous solution of s u g a r (C6H12O6) or any other aqueous solution that is 1.0 m in solute particles b/c i = 1.
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∆ Tb = (1) Kb m The NEW
∆ Tb = (0.52°C/m) 1.0m ∆ Tb = 0.52°C higher
boiling point (bpt) = normal bpt + ∆ Tb boiling point (bpt) = 100°C + 0.52°C = 100.52°C
It is important to realize that the boilingpoint elevation is proportional to the number of solute particles present in a given quantity of solution. When NaCl dissolves in water, 2 mol of solute particles (1 mol of Na+ and 1 mol of Cl-) are formed for each mole of NaCl that dissolves.
Salt Example(IONIC)
Therefore, a 1 m solution of NaCl in water causes a boiling-point elevation twice (i = 2) as large as a 1 m solution of a nonelectrolyte like sugar. Therefore, a 1.0 m solution of NaCl (salt, ionic, electrolyte) in water causes a
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- 17 - -
boiling-point elevation twice as large as a 1.0 m solution of a nonelectrolyte (molecular compound) such as sucrose. 2 particles b/c Na , Cl The ∆ T = (0.52°C/m) 1.0m x2 +
-
b
NEW
∆ Tb = 1.04°C higher for 1.0m NaCl
boiling point (bpt) = normal bpt + ∆ Tb boiling point (bpt) = 100°C + 1.04°C = 101.04°C
Because NaCl dissociates in water to produce TWO particles, and any strong electrolyte can dissociates into more than one particle; especially, salts (ionic compounds). The intermolecular forces are enhanced; especially, with regard to Ion-dipole forces
EX: List the following aqueous solutions in order of their expected boiling points(in water): 0.050 m CaCl2; 0.15 m NaCl; 0.10 m C12H22O11, a sugar. SOLUTION: First notice that CaCl2, NaCl are strong electrolytes, and C12H22O11 is a nonelectrolyte. The molality total particles or CONCENTRATION of particles is as follows: 0.050 m CaCl2 have i = 3 particles
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= (0.15 m in particles)
-
0.15 m NaCl have i =2 particles =(0.30 m in particles) 0.10 m C12H22O11 have i =1 particle =(0.10 m in particles) In comparing the three solutions (same solvent water; therefore same Kb). The boiling points depend on the TOTAL molality of particles (i *m) in solution, the expected ordering is: 0.15 m NaCl (highest boiling point), 0.050 m CaCl2, 0.10 m C12H22O11 (lowest boiling point). EX:Which of the following solutes will produce the largest decrease in freezing point upon addition of 1 kg of water: 1 mol Co(NO3)2, 2 mol of KCl, or 3 mol of ethylene glycol, C2H6O2? SOLUTION:
Re: m = mol/kg of solvent
Co(NO3)2:1 mol/1Kg = 1m # of particles = Co + 2NO3 = i =3 particles So, Co(NO3)2 is 3(1m) or 3m in particles.
KCl: 2 mol/1Kg = 2m
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# of particles = K + Cl = 2 particles So, KCl is 2(2m) or 4m in particles. C2H6O2 : 3 mol/1Kg = 3m # of particles = C2H6O2 (molecular) = 1 particles So, C2H6O2 is 1(3m) or 3m in particles. Answer: 2 mol of KCl b/c 4m in particles
Summary of Colligative Properties: A solution boils when the vapor pressure of the liquid = the external pressure (pushing down on the liquid) Lowering the vapor pressure: Lowered by adding a nonvolatile solute. Decrease the mole fraction of the SOLVENT (means add more solute) results in an elevated boiling point for the SOLUTION:
PA = XA P°A
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Boiling Point Elevation/Freezing point lowering:
∆Tb = iKb m or ∆Tf = iKf m Dramatically affected by increasing the overall molality of particles in solution (i*m). Intermolecular forces exherting an affect!
-
Properties of Solutions
-2--
The Solution Process A solution is a ho m o g e n e o u s m i x t u r e of solute (present in smallest amount) and solvent (present in largest amount).
→ Solutes and solvents are THE components of the solution.
→ A SATURATED SOLUTION contains the maximum amount of solute.
→ Solubility depends on intermolecular forces. In the process of making solutions, intermolecular forces become rearranged.
Consider NaCl (solute) dissolving in water (solvent):
→ The water H-bonds have to be interrupted, → NaCl dissociates into Na+(aq) and Cl-(aq), → io n -di p o l e forces form, → We say the ions are
-
-3--
Factors Affecting Solubility Solute-Solvent Interactions
Polar liquids tend to dissolve in polar solvents, (LIKE DISSOLVES LIKE).
Miscible liquids: MIX in any proportions. Immiscible liquids: DO NOT MIX. Intermolecular forces are important: water and ethanol are miscible because hydrogen bonding in both pure liquids are re-established in the mixture.
Fa c t o r s to consider in deter mining solubility in various SOLVENTS
1. The NUMBER of carbon atoms in a chain affect solubility: the m o r e C atoms, the LESS soluble in water.
-
-4--
-
2. The number of -OH groups within a molecule inc r e a s e s solubility in water (via hydrogen bonding).
Structure of glucose. Note that OH groupS capable of hydrogen bonding with water are prominent on the "surface" of the molecule.
-5--
-
3. The m o r e po l a r bo n d s in the molecule, the better it dissolves in a polar solvent. 4. The le s s po l a r t h e m o l e c u l e the less it dissolves in a polar solvent and the
better is dissolves in a non-polar solvent.
→
Generalization: "like dissolves like".
EX: Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride, CCl4, which is nonpolar, or in water, which is polar: C7H16, NaHCO3, HCl, I2. SOLUTION:
→ →
Both C7H16 and I2 are nonpolar. We would therefore predict that they would be more soluble in CCl4 than in H2O. NaHCO3 is ionic, and HCl is polar covalent. Water would be a better solvent than CCl4 for these two substances.
-6--
-
Naphthalene, sometimes used as moth crystals, has the chemical formula C10H8. It is a ring structure that looks a lot like two hexagonal benzene rings sharing a common two-carbon side. In which of the following solvents will naphthalene dissolve the best?
a) H2O(l) b) CH3OH(l) (methanol) c) C6H6(l) (benzene) d) NH3(l) e) CH3CH2OH(l) (ethanol)
-7--
Factors Affecting Solubility Solubility of gas depends on Pressure Effects
Solubility of a gas in a liquid increases in direct proportion to its partial pressure (relative to the gases in the atmosphere or surroundings) above the solution.
The hi g h e r t h e pr e s s u r e s, the more molecules of gas are clo s e to the solvent and the greater the chance of a gas molecule striking the surface and entering the solution. Therefore, the higher the pressure, the greater the solubility.
The lower the pressure, the fewer m olecules of gas are close to the solvent, and the lower the solubility.
-
-8--
Henry’s Law The relationship between PRESSURE and the SOLUBILITY of a GAS is expressed by the following simple equation:
Sg = kPg Sg is the solubility of the gas (mol/L) k is Henry’s law constant (mol/L-atm), which is unique for each solute-solvent pair
Pg is the partial pressure of the gas above the solvent
Example: Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over liquid at 25°C. The Henry’s Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm. Sg = kPg = (3.1 x 10-2 mol/L-atm)( 4.0 atm) Sg = 0.12 mol/L or 0.12 M
Temperature Effects on Solubility
-
-9--
→ → → → →
Experience tells us that sugar dissolves better in warm water than cold. As t e m p e r a t u r e inc r e a s e s , solubility of solids generally increases. Sometimes (rarely), solubility decreases as tem perature increases (e.g., Ce 2 (SO 4 )3 ). Experience tells us carbonated beverages go flat as they get warm.
Thermal pollution: if lakes get too warm, CO2 and O2 become les s sol u b l e and are not available for plants or animals.
Mole Fraction, Molarity, and Molality
-
- 10 - -
We often use concentration expressions based on the number of moles of one or more components of the solution.
The three we use most commonly are mole fraction, Molarity, and molality.
We introduced Molarity (M) in chapter 4. Recall that the Molarity of a solute in a solution is defined as:
Recall from chapter 10, that the mole fraction of a component of a solution is given by:
or
→
Xsolute = nsolute/ nsolute + nsolvent
The symbol, X, is commonly used for mole fraction, with a subscript to indicate the component of interest.
→ NEW! The MOLALITY of a solution, denoted m, is defined as the number of moles of solute per kil o g r a m (DON’T FORGET) of SOLVENT:
-
- 11 - -
EX: A solution of hydrochloric acid contains 36 grams of HCl and 64 grams of water. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution. SOLUTION (a) The solution contains 36 g of HCl and 64 g of H2O. To calculate the mole fraction of HCl, XHCl, we first convert mass to moles: mol’s of HCl = 36g/36.5 gmol-1 = 0.986 mol mol’s of H2O = 64g/18 gmol-1 = 3.56 mol XHCl = 0.986/0.986 + 3.56 = 0.22 (b) To calculate the molality of HCl in the solution, we use water is the solvent. We calculated the number of moles of HCl in part (a), and the mass of solvent is 64 g = 0.064 kg. Don’t forget to convert!! m of HCl = 0.986 mol/0.064 kg = 15.4 m
Colligative Properties What is it?
FREEZING POINT LOWERING and BOILING POINT ELEVATION are physical properties of solutions that depend on the quantity (molarity or molality) of molecules, but not the kind or identity of solute particles.
-
- 12 - -
-
One consequence of these colligate properties results from: LOWERING VAPOR PRESSURE
Vapor pressure in liquids goes do w n w he n nonvolatile (don’t boil off easily)solutes are dissolved in the m .
Nonvolatile contaminants (ie. Solute) make liquids harder to boil and freeze [EX: antifreeze/water mixtures],
The extent to which a nonvolatile solute lowers its vapor pressure (harder to boil - less vaporneed more heat) is proportional to its concentration. The relationship expressed by Raoult’s Law.
Raoult’s Law := states that the partial
pressure exerted by a solvent vapor above a solution, PA, equals the product of the mole fraction of the solvent in the solution, XA, times the vapor pressure of the pure solvent, P°A.
P A = X A P °A
Example: The vapor pressure of water is 17.5 torr at 20°C. At constant temperature you add
- 13 - -
sugar to water so that the resulting soltution has a 0.8 mol of water and 0.4 mol of sugar. Then the vapor pressure of water the solvent should be lowered making it boil at a higher temperature. Calculate: We are interested in the vapor pressure of the solvent, water, need XH2O: XH2O = nH2O/ nsugar+ nwater = 0.8/0.4 + 0.8 = 0.67 PA = XA P°A = PH2O = 0.67(17.5 torr) = 11.725 torr The presence of the nonvolatile substance (sugar) lowered the vapor pressure of the solvent 5.8 torr! Requiring more heat to boil the solution!
Also, bear in mind from chapter 11, that intermolecular forces are at play! (H-bonding in the above example)
-
- 14 - -
-
Te m p e r a t u r e shi f t ,∆T, of mp or bp from a NORMAL value is proportional to m ole fraction of the conta minant dissolved in an nonvolatile solution. (boiling point is increased or freezing point lowered)
decreased
increased
The INCREASE in boi l i n g poi n t relative to that of the pure solvent, ∆Tb, is directly proportional to the number of solute particles per mole of solvent molecules.
∆Tb is proportional to molality: ∆Tb = iKb m i = van't Hoff factor For molecular solutes, i = 1 For ionic solutes, i = total number of cations and anions (ex., i = 3 for MgCl2 )
- 15 - -
Kb := is the molal boiling point elevation constant
-
(see table below)
m := molality expresses the number of moles of solute per 1000 g of solvent.
Like the boiling-point elevation, the DECREASE in freezing point, ∆Tf, is directly proportional to the molality of the solute: ∆ Tf = iKf m
New Boiling pt of soln. = normal bp + ∆ Tb (increased) New freezing pt of soln. = normal fp - ∆ Tf (decreased)
Sugar Example (Molecular, I =1): For water, Kb is 0.52°C/m; therefore, a 1.0 m aqueous solution of s u g a r (C6H12O6) or any other aqueous solution that is 1.0 m in solute particles b/c i = 1.
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∆ Tb = (1) Kb m The NEW
∆ Tb = (0.52°C/m) 1.0m ∆ Tb = 0.52°C higher
boiling point (bpt) = normal bpt + ∆ Tb boiling point (bpt) = 100°C + 0.52°C = 100.52°C
It is important to realize that the boilingpoint elevation is proportional to the number of solute particles present in a given quantity of solution. When NaCl dissolves in water, 2 mol of solute particles (1 mol of Na+ and 1 mol of Cl-) are formed for each mole of NaCl that dissolves.
Salt Example(IONIC)
Therefore, a 1 m solution of NaCl in water causes a boiling-point elevation twice (i = 2) as large as a 1 m solution of a nonelectrolyte like sugar. Therefore, a 1.0 m solution of NaCl (salt, ionic, electrolyte) in water causes a
-
- 17 - -
boiling-point elevation twice as large as a 1.0 m solution of a nonelectrolyte (molecular compound) such as sucrose. 2 particles b/c Na , Cl The ∆ T = (0.52°C/m) 1.0m x2 +
-
b
NEW
∆ Tb = 1.04°C higher for 1.0m NaCl
boiling point (bpt) = normal bpt + ∆ Tb boiling point (bpt) = 100°C + 1.04°C = 101.04°C
Because NaCl dissociates in water to produce TWO particles, and any strong electrolyte can dissociates into more than one particle; especially, salts (ionic compounds). The intermolecular forces are enhanced; especially, with regard to Ion-dipole forces
EX: List the following aqueous solutions in order of their expected boiling points(in water): 0.050 m CaCl2; 0.15 m NaCl; 0.10 m C12H22O11, a sugar. SOLUTION: First notice that CaCl2, NaCl are strong electrolytes, and C12H22O11 is a nonelectrolyte. The molality total particles or CONCENTRATION of particles is as follows: 0.050 m CaCl2 have i = 3 particles
-
- 18 - -
= (0.15 m in particles)
-
0.15 m NaCl have i =2 particles =(0.30 m in particles) 0.10 m C12H22O11 have i =1 particle =(0.10 m in particles) In comparing the three solutions (same solvent water; therefore same Kb). The boiling points depend on the TOTAL molality of particles (i *m) in solution, the expected ordering is: 0.15 m NaCl (highest boiling point), 0.050 m CaCl2, 0.10 m C12H22O11 (lowest boiling point). EX:Which of the following solutes will produce the largest decrease in freezing point upon addition of 1 kg of water: 1 mol Co(NO3)2, 2 mol of KCl, or 3 mol of ethylene glycol, C2H6O2? SOLUTION:
Re: m = mol/kg of solvent
Co(NO3)2:1 mol/1Kg = 1m # of particles = Co + 2NO3 = i =3 particles So, Co(NO3)2 is 3(1m) or 3m in particles.
KCl: 2 mol/1Kg = 2m
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-
# of particles = K + Cl = 2 particles So, KCl is 2(2m) or 4m in particles. C2H6O2 : 3 mol/1Kg = 3m # of particles = C2H6O2 (molecular) = 1 particles So, C2H6O2 is 1(3m) or 3m in particles. Answer: 2 mol of KCl b/c 4m in particles
Summary of Colligative Properties: A solution boils when the vapor pressure of the liquid = the external pressure (pushing down on the liquid) Lowering the vapor pressure: Lowered by adding a nonvolatile solute. Decrease the mole fraction of the SOLVENT (means add more solute) results in an elevated boiling point for the SOLUTION:
PA = XA P°A
- 20 - -
Boiling Point Elevation/Freezing point lowering:
∆Tb = iKb m or ∆Tf = iKf m Dramatically affected by increasing the overall molality of particles in solution (i*m). Intermolecular forces exherting an affect!
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