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January 11, 2011
December 28, 2010
Hey, Check This Out! A Regional Science High School III Blog.
are those Properties of Solutions that depend on the Number of Dissolved Particles in Solution, but NOT on the Identities of the Solutes.
vapor pressure lowering
The escaping tendency of a Solvent is measured by its Vapor Pressure, which is dependent on the Temperature.
· Vapor Pressure measures the concentration of solvent molecules in the Gas phase.
· Assuming the solute is nonvolatile, the only particles in the gas phase are Solvent Molecules.
· In a solution, fewer solvent molecules are at the surface compared to the pure solvent, so a smaller proportion of solvent molecules will be in the gas phase and the Vapor Pressure for the Solution is lower than that for the Pure Solvent.
Raoult's Law states that for an Ideal Solution the partial Vapor Pressure of a component in Solution is Equal to the Mole Fraction of that component times its Vapor Pressure when pure:
Pa = XaPao
Pa = Vapor Pressure of the Solution
Pao = Vapor Pressure of Pure Solvent
Xa = Mole Fraction of the Solvent
Pa = Vapor Pressure of the Solution
Pao = Vapor Pressure of Pure Solvent
Xa = Mole Fraction of the Solvent
· The Fractional Vapor Pressure Lowering is Equal to the Mole Fraction of the solute:
Xb = (Pao - Pa) ÷ Pao
Pa = Vapor Pressure of the Solution
Pao = Vapor Pressure of Pure Solvent
Xb = Mole Fraction of the Solute
Pa = Vapor Pressure of the Solution
Pao = Vapor Pressure of Pure Solvent
Xb = Mole Fraction of the Solute
Example: Calculating the Vapor Pressure of a Solvent
1.00g of nonvolatile sulfanilamide, C6H8O2N2S, is dissolved in 10.0g of acetone, C3H6O.
The vapor pressure of pure acetone at the same temperature is 400 mmHg.
Calculate the vapor pressure of the solution.
The vapor pressure of pure acetone at the same temperature is 400 mmHg.
Calculate the vapor pressure of the solution.
- Calculate moles of solute: n (C6H8O2N2S) = mass ÷ MM
n(C6H8O2N2S) = 1.00g ÷ (6 x 12 + 8 x 1 + 2 x 16 + 2 x 14 + 32.1) g/mol
= 1.00 ÷ 172.1 = 0.0058 mol - Calculate moles of solvent: n(C3H6O) = mass ÷ MM
n(C3H6O) = 10.0g ÷ (3 x 12 + 6 x 1 + 16) g/mol
= 10.0 ÷ 58 = 0.172 mol - Calculate the mole fraction of the solvent: Xsolvent = nsolvent ÷ (nsolute + nsolvent)
Xa = n(C3H6O) ÷ [n(C3H6O) + n(C6H8O2N2S)]
= 0.172 ÷ [0.172 + 0.0058] = 0.967 - Calculate the vapor pressure: Pa = XaPao
Pa = 0.967 x 400 mmHg = 386.8 mmHg = 387 mmHg
Example: Calculating the Molecular Mass of a Solute
5.00g of a nonvolatile compound was dissolved in 100g of water at 30oC.
The vapor pressure of the solution was measured and found to be 31.20 Torr.
The vapor pressure of pure water at 30oC is 31.82 Torr.
Calculate the molecular mass (formula weight) of the unknown solute.
The vapor pressure of the solution was measured and found to be 31.20 Torr.
The vapor pressure of pure water at 30oC is 31.82 Torr.
Calculate the molecular mass (formula weight) of the unknown solute.
- Calculate the mole fraction of the solute: Xb = (Pao - Pa) ÷ Pao
Xb = (31.82 - 31.2) ÷ 31.82 = 0.0195 - Calculate moles of solvent: n(H2O) = mass ÷ molecular mass
n(H2O) = 100g ÷ (2 x 1 + 16)g/mol = 5.556mol - Calculate moles of solute: Xsolute = nsolute ÷ (nsolute + nsolvent)
0.0195 = nsolute ÷ (nsolute + 5.556)
0.0195(nsolute + 5.556) = nsolute
0.0195(nsolute) + 0.108 = nsolute
0.108 = nsolute - 0.0195(nsolute)
0.108 = 0.9805(nsolute)
(nsolute) = 0.108 ÷ 0.9805 = 0.11 mol - Calculate molecular mass of solute: MM = mass ÷ moles
MM = 5.00g ÷ 0.11 mol = 45.45 g/mol
Microcopic Views:
boiling point elevation
A liquid boils at the temperature at which its vapor pressure equals atmospheric pressure.
The presence of a solute lowers the vapor pressure of the solution at each temperature, making it necessary to heat the solution to a higher temperature to boil the solution.
ΔTb = Kbm
ΔTb = the amount by which the boiling point is raised
m = molality (moles solute particles per kg of solution)
Kb = molal boiling-point elevation constant (solvent dependent)
Boiling Point of solution = normal boiling point of solvent + ΔTb
freezing point depression
A solute lowers the freezing point of a solvent.
In dilute solutions, the freezing point depression is proportional to the molality of the solute particles:
ΔTf = -Kfm
ΔTf = the amount by which the freezing point is lowered
m = molality (moles solute particles per kg of solution)
Kf = molal freezing-point depression constant (solvent dependent)
Freezing Point of solution = normal freezing point of solvent + ΔTf
ΔTb = the amount by which the boiling point is raised
m = molality (moles solute particles per kg of solution)
Kb = molal boiling-point elevation constant (solvent dependent)
Boiling Point of solution = normal boiling point of solvent + ΔTb
A solute lowers the freezing point of a solvent.
In dilute solutions, the freezing point depression is proportional to the molality of the solute particles:
ΔTf = -Kfm
ΔTf = the amount by which the freezing point is lowered
m = molality (moles solute particles per kg of solution)
Kf = molal freezing-point depression constant (solvent dependent)
Freezing Point of solution = normal freezing point of solvent + ΔTf
Some Boiling-Point Elevation and Freezing-Point Depression Constants
Solvent | Boiling Point | Kb | Freezing Point | Kf |
Benzene | 80.2 | 2.53 | 5.5 | 5.12 |
Water | 100.0 | 0.512 | 0.000 | 1.855 |
Acetic Acid | 118.5 | 3.07 | 16.6 | 3.90 |
Camphor | 208.3 | 5.95 | 178.4 | 40.0 |
Naphthalene | 218.0 | 5.65 | 80.2 | 6.9 |
Example: Calculating Boiling and Freezing Point of a Non electrolyte Solution
For a 0.262m solution of sucrose in water, calculate the freezing point and the boiling point of the solution. Freezing Point Calculation:
Calculate the freezing point depression:
ΔTf = -Kf m
Kf = 1.855 (from table above)
m = 0.262m
ΔTf = -1.855 x 0.262 = -0.486oC
ΔTf = -Kf m
Kf = 1.855 (from table above)
m = 0.262m
ΔTf = -1.855 x 0.262 = -0.486oC
Calculate the freezing point of the solution:
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.486
= -0.486oC
Calculate the boiling point of the solution:
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.134
= 100.134oC
Calculate the freezing point of the solution:
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.557
= -0.557oC
Molecules such as solvent molecules that can pass through the membrane will migrate from the side of higher concentration to the side of lower concentration in a process known as osmosis.
The pressure required to stop osmosis is called the osmotic pressure.
In dilute solutions, osmotic pressure (Π) is directly proportional to the molarity of the solution and its temperature in Kelvin.
Van't Hoff Equation: Π = MRT
Π = osmotic pressure
M = molarity = moles ÷ volume(L)
R = ideal gas constant
T = temperature (K)
Solvent can be removed from a solution using a pressure greater than the osmotic pressure. This is known as reverse osmosis.
Example:
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.486
= -0.486oC
Boiling Point Calculation:
Calculate the boiling point elevation:
ΔTb = Kbm
Kb = 0.512 (from table above)
m = 0.262m
ΔTb = 0.512 x 0.262 = 0.134oC
ΔTb = Kbm
Kb = 0.512 (from table above)
m = 0.262m
ΔTb = 0.512 x 0.262 = 0.134oC
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.134
= 100.134oC
Example: Calculating Boiling and Freezing Point of an Electrolyte Solution
Calculate the freezing point and boiling point for a 0.15m aqueous solution of sodium chloride.
Freezing Point Calculation:
Calculate the freezing point depression:
ΔTf = -Kfm
Kf = 1.855 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15m
m(Cl-) = 0.15m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30m
ΔTf = -1.855 x 0.30 = -0.557oC
ΔTf = -Kfm
Kf = 1.855 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15m
m(Cl-) = 0.15m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30m
ΔTf = -1.855 x 0.30 = -0.557oC
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.557
= -0.557oC
Boiling Point Calculation:
Calculate the boiling point elevation:
ΔTb = Kbm
Kb = 0.512 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15m
m(Cl-) = 0.15m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30m
ΔTb = 0.512 x 0.30 = 0.154oC
ΔTb = Kbm
Kb = 0.512 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15m
m(Cl-) = 0.15m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30m
ΔTb = 0.512 x 0.30 = 0.154oC
Calculate the boiling point of the solution:
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.154
= 100.154oC
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.154
= 100.154oC
Videos Related to Boiling Point Elevation and Freezing Point Depression:
TRY THIS!
osmotic pressure
Osmotic pressure arises when two solutions of different concentrations, or a pure solvent and a solution, are separated by a semi permeable membrane.
Molecules such as solvent molecules that can pass through the membrane will migrate from the side of higher concentration to the side of lower concentration in a process known as osmosis.
The pressure required to stop osmosis is called the osmotic pressure.
In dilute solutions, osmotic pressure (Π) is directly proportional to the molarity of the solution and its temperature in Kelvin.
Van't Hoff Equation: Π = MRT
Π = osmotic pressure
M = molarity = moles ÷ volume(L)
R = ideal gas constant
T = temperature (K)
Solvent can be removed from a solution using a pressure greater than the osmotic pressure. This is known as reverse osmosis.
Example:
0.500g hemoglobin was dissolved in enough water to make 100.0mL of solution. At 25oC the osmotic pressure was found to be 1.78 x 10-3 atm. Calculate the molecular mass of the Hemoglobin.
Calculate the molarity, M, of the solution:
M = Π ÷ RT
Π = 1.78 x 10-3 atm
R = 0.0821 L atm K-1mol-1
T = 25oC = 25 + 273 = 298K
M = 1.78 x 10-3 ÷ 0.0821 x 298 = 7.28 x 10-5mol/L
M = Π ÷ RT
Π = 1.78 x 10-3 atm
R = 0.0821 L atm K-1mol-1
T = 25oC = 25 + 273 = 298K
M = 1.78 x 10-3 ÷ 0.0821 x 298 = 7.28 x 10-5mol/L
Calculate the moles, n, of hemoglobin present in solution:
n = M x V
M = 7.28 x 10-5mol/L
V = 100.0mL = 100.0 x 10-3L
n = 7.28 x 10-5 x 100.0 x 10-3 = 7.28 x 10-6mol
n = M x V
M = 7.28 x 10-5mol/L
V = 100.0mL = 100.0 x 10-3L
n = 7.28 x 10-5 x 100.0 x 10-3 = 7.28 x 10-6mol
Calculate the molecular mass, MM, of the hemoglobin:
MM = mass ÷ n
mass = 0.500g
n = 7.28 ÷ 10-6mol
MM = 0.500 ÷ 7.28 x 10-6 = 68,681 g/mol
MM = mass ÷ n
mass = 0.500g
n = 7.28 ÷ 10-6mol
MM = 0.500 ÷ 7.28 x 10-6 = 68,681 g/mol
Videos:
Let's Have Some FUN, This Beat is Sick! Let's Sing the Colligative Properties Song!
Lyrics:
The Amount of Solute in a Solution
Is an attribution
To a Change in Properties
These Properties are all Physical
But that doesn't mean they're Dull
In fact, they're rather Cool
And I said they're called Colligative Properties
There are four of them that I know
Vapor Vressure, Boiling Point
Osmotic Pressure, Freezing Point too!
Vapor Pressure is known to Decrease
With the Concentration Increase
Its the First Property
Its the Pressure from Gas Particles
That have Escaped the Liquid Surface
In a closed space
When you Add Solute to the Solution
Some Liquid Particles Bond to the Solute
So less escape and that decreases the Pressure
Electrolytes have more of an effect
Boiling Point and Freezing Point
Are the next two points
That are affected
The Formula for the Change in Temperature
Is constant times Molality
For every particle
From the boiling point of the pure solvent
You add the change in temperature
For Freezing you Subtract the Change
For more Concentration, there's more Change!
Osmotic pressure, the last property
Is pressure exerted by particles
Of the Solute and Solvent, which is dependent
On the Concentration of the Solute
And I said these are Colligative Properties
Electrolytes have more of an effect
On all of them. They are Properties of Solutions
And I think its an Awesome Subject!
Is an attribution
To a Change in Properties
These Properties are all Physical
But that doesn't mean they're Dull
In fact, they're rather Cool
And I said they're called Colligative Properties
There are four of them that I know
Vapor Vressure, Boiling Point
Osmotic Pressure, Freezing Point too!
Vapor Pressure is known to Decrease
With the Concentration Increase
Its the First Property
Its the Pressure from Gas Particles
That have Escaped the Liquid Surface
In a closed space
When you Add Solute to the Solution
Some Liquid Particles Bond to the Solute
So less escape and that decreases the Pressure
Electrolytes have more of an effect
Boiling Point and Freezing Point
Are the next two points
That are affected
The Formula for the Change in Temperature
Is constant times Molality
For every particle
From the boiling point of the pure solvent
You add the change in temperature
For Freezing you Subtract the Change
For more Concentration, there's more Change!
Osmotic pressure, the last property
Is pressure exerted by particles
Of the Solute and Solvent, which is dependent
On the Concentration of the Solute
And I said these are Colligative Properties
Electrolytes have more of an effect
On all of them. They are Properties of Solutions
And I think its an Awesome Subject!
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