1. Properties of Solution

2. Solution
Defined …
- Homogeneous mixture composed of
only one phase …

3. Solution Defined … Homogeneous mixture composed of only one phase …
In a mixture, a solute is dissolved in
another substance, the solvent…

4. Solution Defined … Homogeneous mixture composed of only one phase …
In a mixture, a solute is dissolved in another substance, the solvent…
The solution takes on the
characteristics of the solvent,
which is the major fraction of the
mixture…

5. Solution Defined … Homogeneous mixture composed of only one phase …
In a mixture, a solute is dissolved in
another substance, the solvent…
The solution takes on the characteristics of the solvent, which is the major fraction of the mixture…
The concentration of a solute in a
solution is a measure of how much
of that solute is dissolved in the
solvent…

6. Homogeneous Solutions
Characteristics:
Ratio of solute to solvent remains the same throughout the solution…

7. Homogeneous Solutions
Characteristics:
Ratio of solute to solvent remains the same
throughout the solution…
Solute will not settle out, no matter how long the solution sits…

8. Homogeneous Solutions
Characteristics:
Ratio of solute to solvent remains
the same throughout the solution…
Solute will not settle out, no matter
how long the solution sits…
Solute and solvent can’t be
separated by filtration or
centrifugation…

9. Homogeneous Solutions
Characteristics:
Ratio of solute to solvent remains the same throughout the solution…
Solute will not settle out, no matter
how long the solution sits…
Solute and solvent can’t be separated
by filtration or centrifugation…
Solution has one phase, i.e., liquid or
gas, such as salt water or air…

10. Solution Composition

11. Solution Composition Temperature Independent Concentration Terms
Mass Percent … Mole Fraction … Molality

12. Mass Percent = (gsolute/gsolvent) x 100%
Solution Composition
Mass Percent = (gsolute/gsolvent) x 100%

13. Mass Percent = (gsolute/gsolvent) x 100%
Solution Composition
Mass Percent = (gsolute/gsolvent) x 100%
Mole Fraction of A (XA) = nA/ntotal = no. of molesA / total no. of moles

14. Solution Composition Mass Percent = (gsolute/gsolvent) x 100%
Mole Fraction of A (XA) = nA/ntotal = no. of molesA / total no. of moles
Molality (m) = nsolute/kgsolvent

15. Solution Composition
A solution is prepared by adding 5.84 gH2CO to gH2O. The final volume of the solution is mL. Calculate mass percent, mole fraction, molality and molarity of the formaldehyde in solution.

16. Solution Composition Known:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO

17. Solution Composition Known:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO
massH2O = gH2O = kgH2O

18. Solution Composition Known:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO
massH2O = gH2O = kgH2O
Vfinal = mL = Lfinal

19. Solution Composition Known:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO
massH2O = gH2O = kgH2CO
gmmH2O = 18.0 gH2O
Vfinal = mL = Lfinal

20. Solution Composition Known: Unknown:
mH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O
gmmH2O = 18.0 gH2O
Vfinal = mL = Lfinal

21. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O
Vfinal = mL = Lfinal

22. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal

23. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO

24. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO

25. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O

26. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO

27. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100%= 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2CO = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2)

28. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2CO = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2O

29. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2CO = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2O
XH2CO = ? || nH2CO / ntotal = (0.194 molH2CO) / (0.194 molH2CO molH2O) =

30. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2CO = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2O
XH2CO = ? || nH2CO / ntotal = (0.194 molH2CO) / (0.194 molH2CO molH2O) =
XH2O = ? || 1.0 – =

31. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2CO = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2O
XH2CO = ? || nH2CO / ntotal = (0.194 molH2CO) / (0.194 molH2CO molH2O) =
XH2O = ? || 1.0 – =

32. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2O = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2O
XH2CO = ? || nH2CO / ntotal = (0.194 molH2CO) / (0.194 molH2CO molH2O) =
XH2O = ? || 1.0 – =
mH2CO = ? mH2CO /KgH2O || nH2CO / KgH2O = molH2CO / KgH2O = 1.94 mH2CO

33. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2O = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2O
XH2CO = ? || nH2CO / ntotal = (0.194 molH2CO) / (0.194 molH2CO molH2O) =
XH2O = ? || 1.0 – =
mH2CO = ? mH2CO /KgH2O || nH2CO / KgH2O = molH2CO / KgH2O = 1.94 mH2CO
M = ? molH2CO/Vfinal || molH2CO / .104 Lfinal = 1.87 MH2CO

34. Solution Composition Known: Unknown:
massH2CO = 5.84 gH2CO ; gmmC2HO = gH2CO mass%H2CO = ? %H2CO
massH2O = gH2O = kgH2O XH2CO = ?
gmmH2O = 18.0 gH2O mH2CO = ? mH2CO
Vfinal = mL = Lfinal molarityH2CO = ? MH2CO
Solution:
Mass%H2CO = ? %H2CO || (gsolute/gsolution) x 100% = (5.84 gH2CO / gH2CO + gH2O) x 100% = 5.52%H2CO
MassH2O = ? %H2O || 100% - %H2O = 100% %H2CO = 94.48%H2O
Preliminary Calculations:
molesH2CO = (5.84 g H2CO) (1 moleH2CO/30.03 gH2CO)) = molH2CO
molesH2O = (100.0 gH2O) x (1 moleH2O/18.02 gH2O) = 5.55 molH2O
XH2CO = ? || nH2CO / ntotal = (0.194 molH2CO) / (0.194 molH2CO molH2O) =
XH2O = ? || 1.0 – =
mH2CO = ? mH2CO /KgH2O || nH2CO / KgH2O = molH2CO / KgH2O = 1.94 mH2CO
M = ? molH2CO/Vfinal || molH2CO / .104 Lfinal = 1.87 MH2CO
Note: As the temperature of a solution changes, the molarity changes due to solution volume changes.

35. Solution Composition
Normality (N) = equivalentssolute / Lsolution

36. Normality (N) = equivalentssolute / Lsolution
Solution Composition
Normality (N) = equivalentssolute / Lsolution
Note:
An equivalent is related to a mole of substance in solution…

37. Normality (N) = equivalentssolute / Lsolution
Solution Composition
Normality (N) = equivalentssolute / Lsolution
Note:
An equivalent is related to a mole of substance in solution…
No.equivalents > No.moles of solute

38. Normality (N) = equivalentssolute / Lsolution
Solution Composition
Normality (N) = equivalentssolute / Lsolution
Note:
An equivalent is related to a mole of substance in solution…
No.equivalents > No.moles of solute
An equivalent is the mass of the solute that can furnish or provide on mole of protons (H+ in acid/base) or one mole of electrons (redox).

40. Normality (N) = equivalentssolute / Lsolution
Solution Composition
Normality (N) = equivalentssolute / Lsolution
Note:
An equivalent is related to a mole of substance in solution…
No.equivalents > No.moles of solute
An equivalent is the mass of the solute that can furnish or provide on mole of protons (H+ in acid/base) or one mole of electrons (redox).
An equivalent mass is the mass of one equivalent of substance.

41. Normality (N) = equivalentssolute / Lsolution
Solution Composition
Normality (N) = equivalentssolute / Lsolution
Note:
An equivalent is related to a mole of substance in solution…
No.equivalents > No.moles of solute
An equivalent is the mass of the solute that can furnish or provide on mole of protons (H+ in acid/base) or one mole of electrons (redox).
An equivalent mass is the mass of one equivalent of substance.
The number of equivalents per mole of solute depends on the reaction involved.

42. H3PO4 + 3NaOH -> PO4-3 + 3H2O + 3Na+
Solution Composition
Consider this reaction ….
H3PO4 + 3NaOH -> PO H2O + 3Na+
Given this reaction …
If gH3PO4 in 800 mLH2O

43. Solution Composition Given this reaction …
Consider this reaction ….
H3PO4 + 3NaOH -> PO H2O + 3Na+
Given this reaction …
If gH3PO4 in 800 mLH2O
Each mole of H3PO4 (97.99 g/mol) supplies 3 molesH+
there are 3 equivalents per mole of H3PO4

44. Solution Composition
Consider this reaction ….
H3PO4 + 3NaOH -> PO H2O + 3Na+
Given this reaction …
If gH3PO4 in 800 mLH2O
Each mole of H3PO4 (97.99 g/mol) supplies 3 molesH+
There are 3 equivalents per mole of H3PO4
The equivalent mass … that can supply one mole of protons …
(97.99 gH3PO4/molH3PO4) / (3 eqH+/molH3PO4) = gH3PO4/eqH+

45. Solution Composition EquivalentsH3PO4 ….
Consider this reaction ….
H3PO4 + 3NaOH -> PO H2O + 3Na+
Given this reaction …
If gH3PO4 in 800 mLH2O
Each mole of H3PO4 (97.99 g/mol) supplies 3 molesH+
There are 3 equivalents per mole of H3PO4
The equivalent mass … that can supply one mole of protons …
(97.99 gH3PO4/molH3PO4) / (3 eqH+/molH3PO4) = gH3PO4/eqH+
EquivalentsH3PO4 ….
EquivalentsH3PO4 = (1 eqH+/32.66 gH3PO4) x gH3PO4 = eqH+

46. Solution Composition Normality …
Consider this reaction ….
H3PO4 + 3NaOH -> PO H2O + 3Na+
Given this reaction …
If gH3PO4 in 800 mLH2O
Each mole of H3PO4 (97.99 g/mol) supplies 3 molesH+
There are 3 equivalents per mole of H3PO4
The equivalent mass … that can supply one mole of protons …
(97.99 gH3PO4/molH3PO4) / (3 eqH+/molH3PO4) = gH3PO4/eqH+
EquivalentsH3PO4 ….
EquivalentsH3PO4 = (1 eqH+/32.66 gH3PO4) x gH3PO4 = eqH+
Normality …
NH3PO4 = ? eqH+/Lsolution || (0.870 eqH+) / (0.800 Lsolution) = 1.09 NH3PO4

47. Solution Composition Molarity …
Consider this reaction ….
H3PO4 + 3NaOH -> PO H2O + 3Na+
Given this reaction …
If gH3PO4 in 800 mLH2O
Each mole of H3PO4 (97.99 g/mol) supplies 3 molesH+
There are 3 equivalents per mole of H3PO4
The equivalent mass … that can supply one mole of protons …
(97.99 gH3PO4/molH3PO4) / (3 eqH+/molH3PO4) = gH3PO4/eqH+
EquivalentsH3PO4 ….
EquivalentsH3PO4 = (1 eqH+/32.66 gH3PO4) x gH3PO4 = eqH+
Normality …
NH3PO4 = ? eqH+/Lsolution || (0.870 eqH+) / (0.800 Lsolution = 1.09 NH3PO4
Molarity …
M = molH3PO4 /LH3PO4 = ((1 mol/97.99 gH3PO4) x (28.42 gH3PO4) )/(0.800 LH2O) = MH3PO4

48. Solution Composition Note: N = 3M Consider this reaction ….
H3PO4 + 3NaOH -> PO H2O + 3Na+
Given this reaction …
If gH3PO4 in 800 mLH2O
Each mole of H3PO4 (97.99 g/mol) supplies 3 molesH+
There are 3 equivalents per mole of H3PO4
The equivalent mass … that can supply one mole of protons …
(97.99 gH3PO4/molH3PO4) / (3 eqH+/molH3PO4) = gH3PO4/eqH+
EquivalentsH3PO4 ….
EquivalentsH3PO4 = (1 eqH+/32.66 gH3PO4) x gH3PO4 = eqH+
Normality …
NH3PO4 = ? eqH+/Lsolution || (0.870 eqH+) / (0.800 Lsolution = 1.09 NH3PO4
Molarity …
M = molH3PO4 /LH3PO4 = ((1 mol/97.99 gH3PO4) x (28.42 gH3PO4) )/(0.800 LH2O) = MH3PO4
Note: N = 3M

49. Energy of Solution Formation
There are three steps that must occur in order for a solution to form.

50. Energy of Solution Formation
There are three steps that must occur in order for a solution to form.
Expansion of the Solute (Endothermic, ∆H1 = +)

51. Energy of Solution Formation
There are three steps that must occur in order for a solution to form.
Expansion of the Solute (Endothermic, ∆H1 = +)
Expansion of the Solvent (Endothermic, ∆H2 = +)

52. Energy of Solution Formation
There are three steps that must occur in order for a solution to form.
Expansion of the Solute (Endothermic, ∆H1 = +)
Expansion of the Solvent (Endothermic, ∆H2 = +)
Combining the Solute and Solvent (Exothermic, ∆H3 = -)

53. Energy of Solution Formation
∆Hsolution = ∆H1 + ∆H2 + ∆H3

54. Energy of Solution Formation
Like Dissolves Like!
∆H ∆H ∆H ∆Hsoln Outcome
Polar Solute/Polar Solvent Large(+) Large(+) Large(-) Small(+) Sol Forms
(Salt)/(Water)

55. Energy of Solution Formation
Oil & Water are not miscible!
∆H ∆H ∆H ∆Hsoln Outcome
Polar Solute/Polar Solvent Large(+) Large(+) Large(-) Small(+) Sol Forms
(Salt)/(Water)
Nonpolar Solute/Polar Solvent Small((+) Large(+) Small(-) Large(+) No Sol Forms
(Oil/Wtaer))

56. Energy of Solution Formation
Like Dissolves Like!
∆H ∆H ∆H ∆Hsoln Outcome
Polar Solute/Polar Solvent Large(+) Large(+) Large(-) Small(+) Sol Forms
(Salt)/(Water)
Nonpolar Solute/Polar Solvent Small((+) Large(+) Small(-) Large(+) No Sol Forms
(Oil/Wtaer))
Nonpolar Solute/Nonpolar Solvent Small(+) Smakk(+) Small(-) Small (+) Sol Forms
(Oil/Gasoline)

57. Energy of Solution Formation
∆H ∆H ∆H ∆Hsoln Outcome
Polar Solute/Polar Solvent Large(+) Large(+) Large(-) Small(+) Sol Forms
(Salt)/(Water)
Nonpolar Solute/Polar Solvent Small((+) Large(+) Small(-) Large(+) No Sol Forms
(Oil/Wtaer))
Nonpolar Solute/Nonpolar Solvent Small(+) Smakk(+) Small(-) Small (+) Sol Forms
(Oil/Gasoline)
Polar Solute/Nonpolar Solvent Large(+) Small(+) Small(-) Large(+) No Sol Forms
(Salt/Oil)

58. Entropy
A measure of the randomness of a system …

59. Entropy as a Driving force for Solution Formation
One factor that favors solution formation is an increase in disorder of a system.
The large increase in disorder that occurs when the solute and solvent mix overcome small energies required for the solution of form.
Process that require large amounts of energy tend not to occur.
∆G = ∆H - T∆S

60. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?

61. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?
Known: Unknown:
DensityCCl4 = 1.4 g/mL How would the liquids interact?
DensityH2O = 1.0 g/mL In what layers would the I2 dissolve?
DensityC6H12 = 0.8 g/mL

62. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?
Known: Unknown:
DensityCCl4 = 1.4 g/mL How would the liquids interact?
DensityH2O = 1.0 g/mL In what layers would the I2 dissolve?
DensityC6H12 = 0.8 g/mL
Solution:
1st: Examine the nature of the molecules…

63. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?
Known: Unknown:
DensityCCl4 = 1.4 g/mL How would the liquids interact?
DensityH2O = 1.0 g/mL In what layers would the I2 dissolve?
DensityC6H12 = 0.8 g/mL
Solution:
1st: Examine the nature of the molecules…
CCl4 … non-polar

64. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?
Known: Unknown:
DensityCCl4 = 1.4 g/mL How would the liquids interact?
DensityH2O = 1.0 g/mL In what layers would the I2 dissolve?
DensityC6H12 = 0.8 g/mL
Solution:
1st: Examine the nature of the molecules…
CCl4 … non-polar
H2O … polar

65. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?
Known: Unknown:
DensityCCl4 = 1.4 g/mL How would the liquids interact?
DensityH2O = 1.0 g/mL In what layers would the I2 dissolve?
DensityC6H12 = 0.8 g/mL
Solution:
1st: Examine the nature of the molecules…
CCl4 … non-polar
H2O … polar
C6H12 … non-polar

66. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?
Known: Unknown:
DensityCCl4 = 1.4 g/mL How would the liquids interact?
DensityH2O = 1.0 g/mL In what layers would the I2 dissolve?
DensityC6H12 = 0.8 g/mL
Solution:
1st: Examine the nature of the molecules…
CCl4 … non-polar
H2O … polar
C6H12 … non-polar
Using Densities…
C6H12
H2O
CCl4

67. Predicting Solubility
When 50 mLCCl4, 50 mLH2O and 50 mLC6H12 are added to a 250 mL graduated cylinder, how would the liquids look in the cylinder? Would a solution form? If solid I2 flakes are added to the system, in which layers, if any, would they dissolve?
Known: Unknown:
DensityCCl4 = 1.4 g/mL How would the liquids interact?
DensityH2O = 1.0 g/mL In what layers would the I2 dissolve?
DensityC6H12 = 0.8 g/mL
Solution:
1st: Examine the nature of the molecules…
CCl4 … non-polar
H2O … polar
C6H12 … non-polar
Using Densities…
C6H12
H2O
CCl4
Iodine is polar and mixes with CCl4 & C6H12

68. Structure & Solubility
Determine which of the following molecules is likely to be water soluble.

69. Structure & Solubility
Determine which of the following molecules is likely to be water soluble.
Water soluble … small molecule … 3 polar bonds

70. Structure & Solubility
Determine which of the following molecules is likely to be water soluble.
Water soluble … small molecule … 3 polar bonds
Insoluble … large molecule

71. Structure & Solubility
Determine which of the following molecules is likely to be water soluble.
Water soluble … small molecule … 3 polar bonds
Insoluble … large molecule
Insoluble … nonpolar molecule

72. Structure & Solubility
Determine which of the following molecules is likely to be water soluble.
Water soluble … small molecule … 3 polar bonds
Insoluble … large molecule
Insoluble … nonpolar molecule
Water soluble … small molecule … 2 polar bonds