Colligative Properties. Ch 12. What is a colligative property?. Only apply to solutions (solvent and a solute) Colligative Property: A property that depends only upon the solute concentration and not on the solute’s identity. Four Important Colligative Properties of Solutions.
Related searches for Colligative Properties
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
On the surface of the pure solvent (shown on the left) there are more solvent molecules at the surface than in the right-hand solution flask. Therefore, it is more likely that solvent molecules escape into the gas phase on the left than on the right. Therefore, the solution should have a lower vapor pressure than the pure solvent.
Adding solutes to boiling liquids (like salt to water) raises the boiling point.
For the Boiling Point Elevation: think of the solute molecules interfering with the liquid’s ability to become a gas. They get in the way… so need more energy to overcome = higher temperature
Adding solutes to freezing liquids (like salt to water) lowers the freezing point.
For Freezing Point Depression: solute molecules get in the way of the formation of crystals. In essence, the connections that need to be made are harder to create, so lower temperature is required.
Osmosis refers to the flow of solvent molecules past a semipermeable membrane that stops the flow of solute molecules only. When a solution and the pure solvent are placed on either side of a semipermeable membrane, it is found that more solvent molecules flow out of the pure solvent side of the membrane than solvent flows into the pure solvent from the solution side. This is because more solvent molecules are at the membrane interface on the solvent side of the membrane than on the solution side. Therefore, it is more likely that a solvent molecule will pass from the solvent side to the solution side than vice versa. That difference in flow rate causes the osmotic pressure of the solution to rise.
If colligative properties depend on the amount of the solute in the solvent, then the equations defining them must include a concentration term, and sure enough, they do.
Three colligative properties set equal to three different concentrations terms times a solvent constant
ΔP vapor pressure ΔT temperature raising or lowering π osmotic pressure
ΔP = χP0
ΔT f = mK f
π = MRT
χ (mole fraction)
ΔT b= mK b
Expressing Concentration concentrations terms times a solvent constant
Let’s start by imagining that we are placing 50.0 g (0.146 mole) of the solute, sugar, in 117 g (6.5 mole) of the solvent, H2O. Density of the solution is 1.34 g/ml.
What is χ (mole fraction) of 50g of sugar in 117 g of water?
χ = molesA
molesA + molesB
χ = moles sugar
moles sugar + moles H2O
χ = 0.146
0.146 + 6.5
= 0.0220 mole fraction sugar
What is m (molality) of 50.0 g of sugar in 117 g of water? concentrations terms times a solvent constant
m = moles A
m = 0.146 moles sugar (solute)
0.117 kg H2O (solvent)
m = 1.25 molal
What is M (molarity) of 50.0 g of sugar in 117 g of water? First we need to find the volume of solution from a density calculation.
V solution = (mass)/(density) = (50 g + 117 g)/(1.34 g/mL) = 125 mL
M = moles sugar (solute)
Volume solution (L)
M = 0.146 moles sugar (solute)
0.125 L (solution)
M = 1.17 Molar
So we have 3 ways to describe 50.0 g of sugar in 117 g of water, each of which is used in a colligative property calculation.
0.022 mole fraction ≡ 1.25 m ≡ 1.17 M
So how big of an effect does a solute concentration have on a colligative property? Let’s use our calculated concentrations to find out…
Colligative property 1: Vapor pressure depression a colligative property? Let’s use our calculated concentrations to find out…
ΔP = P0χ mole fraction which is the amount of solute added
Vapor pressure depression constant which is the vapor pressure of pure solvent at a given T. of pure solvent
Ex: For H2O at 25oC the pure vapor pressure is 23.76 mm Hg (see table 5.3 p. 200 for vapor pressure constants)
So the vapor pressure depression in
ΔP = 23.76 mm Hg (0.0220) = 0.523 mm Hg
And the new vapor pressure is now about 23.2 mm Hg.
By the way, this equation is referred to as Raoult’s Law which says simply that the vapor pressure above a solution is proportional to the mole fraction of the solute.
ΔTb= K bm
molality which is the amount of solute added
because of solute (b.p. of
solution minus b.p. of solvent)
temperature constant which is the boiling point elevation constant (see table on p. 532)
For water, Kbis 0.512 oC/molal
So the boiling point elevation for our previous example is ΔTb= Kbm = (0.512 oC/molal)(1.25 molal) = 0.64 oC
And the new b.p. of water with a heck of a lot of sugar in it is 100.64 oC
ΔTf = Kf m
molality which is the amount of solute added
because of solute (f.p. of
solvent minus f.p. of solution)
constant which is the freezing point depression constant (see table p. 532)
For water, Kb is 1.86 oC/molal
So the freezing point depression for our previous example is
ΔT f= Kfm = (1.86 oC/molal)(1.25 molal) = 2.32 oC
And the new f.p. of water with a heck of a lot of sugar in it is
π = MRT
system temperature (K)
increase because of solute
Gas constant, R
molarity which is the amount of solute added
R is 0.0821 L·atm/K·mol
So the osmotic pressure change for our previous example is
π = MRT = (1.17 M)(0.0821 L·atm/K·mol)(298 K) = 28.6 atm
Very simply, Van’t Hoff corrects for the fact that the number of particles you throw into solution is not always the number of particles that determine the magnitude of the property. For example, think about what happens when you put the following one mole quantities into a liter of water. Which one raises the boiling point the most?
1. 1 mole NaCl 2. 1 mole Na2S 3. 1 mole CaS 4. 1 mole sugar
1 mole Na+
and 1 mole
2 moles Na+
1 mole S2−
∆Tf = i Kfm ΔP = i P0χ
∆Tb = i Kbm π = i MRT
What is the freezing point of a solution of 15.0 g of NaCl in 250 g of water? The molal freezing point constant, Kf, for water is 1.86 oC / molal.
For NaCl, i = 2 moles = 15.0 g = 0.257 mole
Kf =1.86 oC / molal. 58.4g/mol
molal = 0.257 mole/ 0.250 kg = 1.03 m in NaCl
∆Tf = i Kfm
∆Tf = (2)(1.86 oC/molal)(1.03 molal) = 3.8 oC. The freezing point of the solution is, therefore, 0 oC - 3.8 oC = -3.8 oC