1. Expressing Solution Composition
Mass fraction =
Mass Solute
Total Mass of Solution
Weight percent =
Mass fraction x 100%
Example
Saline solutions, NaCl(aq), are often used in medicine. What is the weight percent of NaCl in a solution of 4.6 g of NaCl in 500. g of water?
4.6 g
500. g g
mass fraction = =
weight percent = x 100% = 0.91 %
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2. Parts per Million, Billion & Trillion
mass solute
mass solution
% = Parts per hundred =
x 102
Very dilute = low mass fraction.
Units for high dilution (trace solute):
x 106
Parts per million (ppm) =
mass solute
mass solution
x 109
Parts per billion (ppb) =
mass solute
mass solution
x 1012
Parts per trillion (ppt) =
mass solute
mass solution
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3. Parts per Million, Billion & Trillion
1 ppm of solute in water = 1 mg / 1000 g of solution
Since 1 L of water has a mass ≈ 1000 g.
1 ppm ≈ 1 mg/L
Similarly 1 ppb ≈ 1 μg/L
1 ppt ≈ 1 ng/L
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4. Parts per Million, Billion & Trillion
Sea water is 10,600 ppm Na+. Calculate the mass fraction and mass percent of sodium ions in sea water.
Mass fraction
10,600 ppm = 10,600 g Na+ in 106 g of solution
Mass fraction = ,600 g Na+
106 g of solution =
Mass percent = 1.06 %
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5. Molarity Molarity = M = = moles of solute liters of solution n V
Because V varies with T, molarity is T dependent!
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6. Molality Molality is another concentration scale: moles of solute
kilograms of solvent
Molality is another concentration scale:
A mass-based unit.
Uses solvent mass (not solution).
It is T independent (unlike molarity).
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7. Molarity and Molality
Calculate the molarity and molality of an 93.0 % by mass sulfuric acid solution. Its density = g/mL.
93 % = 93.0 g H2SO g H2O (in g soln.)
nH2SO4 =
93.0 g
98.09 g/mol
= mol
Vsoln =
100.0 g
1.835 g/mL
= mL
[H2SO4] =
mol L
= 17.4 M
mol kg
= 1.4 x 102 mol/kg (m)
mH2SO4 =
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8. Mole fraction is the moles of component over the total moles of solution. It is abbreviated as X.
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9. Mole Fraction
Determine the mole fractions of glucose and water in a solution containing 5.67 g of glucose dissolved in 25.2 g of water.
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10. Mole Fraction 5.67 g of glucose = 0.0315 mol of glucose.
Calculate the total moles of solution.
The total moles of solution are:
Therefore,
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11. Colligative Properties of Solutions
Colligative properties are those properties that:
Only depend upon the number of “particles” in solution.
each ion or molecules has the same effect; the particle type is unimportant.
Solvent vapor P drops if non-volatile solute is added.
Lower purity solvent = lower vapor P.
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12. Vapor Pressure Lowering
over pure solvent
Raoult’s law: Psolvent = Xsolvent P°solvent
over the solution
mole fraction
moles of n1
total moles (n1+n2+…)
X1 = =
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13. Raoult’s law
To obtain an explicit expression for the vapor-pressure lowering of a solvent in a solution, assume that Raoult’s law holds and that the solute is a nonvolatile electrolyte. Therefore,
Substituting Raoult’s law gives:
A: solvent
B: nonelectrolyte solute
P = PAXB
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14. Vapor Pressure Lowering
A solution of urea in water has a vapor P of torr. The vapor P of pure water is torr. Calculate the mole fraction of urea in the solution.
Urea is non-volatile. Water obeys Raoult’s law:
Pwater = XwaterP°water
291.2 torr = Xwater(355.1 torr)
Xwater= 291.2/355.1 = 0.820
Xurea = 1 – = 0.180
(Remember: X1 + X2 + … = 1)
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15. Raoult’s Law for Volatile Solute
When both the solvent and the solute can evaporate, both molecules will be found in the vapor phase.
The total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent.
For an ideal solution
Ptotal = Psolute + Psolvent
The solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s.
Psolute = χsolute∙P°solute and Psolvent = χsolvent ∙P°solvent
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16. Such solutions occur when the substances are chemically similar.
An ideal solution for substances A and B is one in which both substances follow Raoult’s law for all values of mole fractions.
Such solutions occur when the substances are chemically similar.
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17. Vapor-Pressure Lowering of Solutions: Raoult’s Law
Ptotal = PA + PB = (PA XA) + (PB XB)
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18. Vapor Pressure of a Nonideal Solution
When the solute–solvent interactions are stronger than the solute–solute  solvent–solvent, the total vapor pressure of the solution will be less than predicted by Raoult’s law, because the vapor pressures of the solute and solvent are lower than ideal.
When the solute–solvent interactions are weaker than the solute–solute  solvent–solvent, the total vapor pressure of the solution will be more than predicted by Raoult’s law.
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19. Deviations from Raoult’s Law
Benzene- toluene
Acetone- water
Ethanol- hexane
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20. Boiling Point Elevation
Non-volatile solutes increase the b.p. of a solvent. Why?
Solvent vapor P is lowered.
Higher T is needed to get the vapor P = external P.
Quantitatively ΔTb = Kb msolute
Note: the Kb value only depends on the solvent,
the molality, m depends on the solute.
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21. Freezing-Point Lowering
A non-volatile solute lowers the f.p. of a solvent:
Quantitatively ΔTf = Kf msolute
Kf is a constant for the solvent.
Ethylene glycol (antifreeze) is added to car radiators
to lower the freezing point of water (stops freezing).
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22. Boiling-Point Elevation and Freezing-Point Depression of Solutions
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23. Boiling-Point Elevation and Freezing-Point Depression of Solutions
Nonelectrolytes
Electrolytes
ΔTb = Kbm
ΔTb = Kbmi
ΔTf = Kfm
ΔTf = Kfmi
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24. Freezing Point Lowering
Calculate the f.p. of an aqueous 30.0% ethylene glycol mixture. For water Kf = -1.86°C kg mol-1.
100.0 g of 30% mix: g C2H2(OH) g H2O
nglycol = 30.0 g / g mol-1 = mol
mglycol = ( mol / kg) = molal
ΔTf = -1.86°C kg mol-1(6.904 mol/kg) = °C
Freezing point = 0.00°C – 12.8°C = -12.8°C
normal freezing point of water
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25. Molar Mass By Boiling Point Elevation
Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is C. What is the molar mass of benzyl acetate?
Given:
bp (chloroform) = C
Kb (chloroform) = 3.63C/m
Solution: The molality of the solution can be found from the freezing point depression.
Knowing molality and mass of solvent, one can find the moles of solute.
Knowing mass and moles of solute, the molar mass of the compound can be determined.
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26. Molar Mass By Boiling Point Elevation, continued
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27. Osmotic Pressure of Solutions
Semipermeable membrane
Allows passage of small “particles”.
Stops large “particles”.
Ion + H2O coordination sphere – too large to pass through
net solvent flow
Large molecule cannot pass.
e.g. animal bladders and cell membranes.
Osmosis
Movement of solvent through a semipermeable membrane from dilute to more concentrated solution.
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28. An Osmosis Cell
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29. Osmosis at the Particulate Level
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30. Osmosis
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31. Osmotic Pressure of Solutions
Osmotic pressure = P that must be applied to stop osmosis.
Π = c R T
gas constant
molarity of solution
absolute T
R = (atm∙L)/(mol∙K)
Height of the column of solution is a measure of Π.
Pure water
Water enters the bag, increasing the P…
semipermeable bag of 5% sugar water
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32. Osmotic Pressure Π = c R T n R T P = = c R T V
Easy to remember. Very similar to the ideal gas law:
c R T
n R T V
P = =
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33. Osmotic Pressure A cell can be exposed to 3 kinds of solution:
Isotonic
[solute]out=[solute]in
No Net flow.
Hypertonic
[solute]out >[solute]in
Net flow out.
Hypotonic
[solute]out <[solute]in
Net flow in.
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34. Reverse Osmosis Used to purify water: Normal Osmosis Reverse Osmosis
Water molecules cross the membrane diluting the brine.
Flow stops when P = Π is applied.
Reverse Osmosis
Flow is reversed if P > Π is applied.
Pure water can be separated from brine.
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35. Example
A solution made up of 2.00 g of an unknown solute and 275 mL of water (d = 1.00 g/mL) has an osmotic pressure of atm at 25°C
What is the molar mass of the unknown solute?
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36. Example Analysis: Information given Mass of solute (2.00 g)
Volume of solvent
(275 mL)
π (0.872 atm); T (25°C)
Density of solvent
(1.00 g/mL)
Density of solution
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37. Example Strategy: Find the molarity M, by substituting in
Determine the volume of the solution by first finding its mass
Mass of solution = mass of solute + mass of water
Volume of solution = mass/density
Find the moles of solute using the defining equation for molarity
M = moles solute/volume of solution (L)
Find the molar mass
Molar mass = mass/moles
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38. Example
Solution: M
Volume of solution
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39. Example
Moles of solute
Molar mass
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40. Colligative Properties of Electrolytes
Ionic solutes dissolve to form more than one particle per formula unit. We alter the colligative property equations to account for this fact by including i, the number of ions per formula unit:
DP = iPAXB
DTb = iKbm
DTf = iKfm
p = iMRT
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41. Colligative Properties of Electrolytes
In aqueous solution:
1 mol sucrose → 1 mol particles
1 mol NaCl → 2 mol particles (Na+ and Cl-)
1 mol CaCl2 → 3 mol (Ca2+ and 2 Cl-)
Modify the formulas by replacing
msolute with isolutemsolute
where i = the number of particles per formula unit.
= van’t Hoff factor
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42. Colligative Properties of Electrolytes
A simple correction, but it only works at low [ion].
Ions attract each other in solution.
Only act as separate units at very low [ion].
i is smaller than expected in many solutions:
[MgSO4]
iexpected
iobserved M 2
2.00
0.005 M
1.72
0.50 M
1.07
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43. Colligative Properties of Strong Electrolyte Solutions
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44. Boiling Point Elevation
At what T will molal NaCl(aq) boil? Kb = 0.51°C kg mol-1
The b.p. of molal NaCl:
ΔTb = Kb isolutemsolute
= 0.51°C kg mol-1(2)(0.100 mol/kg) = 0.10 °C
b.p. = °C °C = °C
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