1. Properties of Solutions
Lesson 2

2. Measuring concentration
Concentration = amount of solute/amount of solution
Weight/volume percent
Mass solute in g/volume of soln in mL x 100%
Weight/weight percent
Mass solute in g/mass solution in g x 100%

3. Moles of solute/Liters of Solution (M) Molality =
Concentration
Molarity =
Moles of solute/Liters of Solution (M)
Molality =
Moles of solute/Kg of Solvent (m)
Mole Fraction=
Moles solute/total number of moles
Mass %=
Mass solute/total mass x 100
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4. Moles of solute/Liters of Solution (M) Molality =
Concentration
Molarity =
Moles of solute/Liters of Solution (M)
Molality =
Moles of solute/Kg of Solvent (m)
Mole Fraction=
Moles solute/total number of moles
Mass %=
Mass solute/total mass x 100
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5. Moles of solute/Liters of Solution (M) Molality =
Concentration
Molarity =
Moles of solute/Liters of Solution (M)
Molality =
Moles of solute/Kg of Solvent (m)
Mole Fraction=
Moles solute/total number of moles
Mass %=
Mass solute/total mass x 100
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6. Molarity

7. Convert the given grams of solute to moles of solute :
A sample of NaNO3 weighing 8.5 grams is placed in a 500 ml volumetric flask and distilled water was added to the mark on the neck of the flask. Calculate the Molarity of the resulting solution.
Convert the given grams of solute to moles of solute :
Convert given ml of solution to liters
Apply the definition for Molarity: Molarity = moles NaNO3 / volume of the solution in liters
M = 0.1 mole / .500 liters = Molar NaNO3
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8. Exercise #1
You have 1.00 mol of sugar in mL of solution. Calculate the concentration in units of molarity.
8.00 M

9. Exercise #2
You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?
0.200 L

10. Exercise #3
Consider separate solutions of NaOH and KCl made by dissolving g of each solute in mL of solution. Calculate the concentration of each solution in units of molarity.
10.0 M NaOH
5.37 M KCl

11. Moles of solute/Liters of Solution (M) Molality =
Molarity =
Moles of solute/Liters of Solution (M)
Molality =
Moles of solute/Kg of Solvent (m)
Mole Fraction=
Moles solute/total number of moles
Mass %=
Mass solute/total mass x 100
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12. Molality

13. Exercise #4
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)
0.816 m

14. Moles of solute/Liters of Solution (M) Molality =
Molarity =
Moles of solute/Liters of Solution (M)
Molality =
Moles of solute/Kg of Solvent (m)
Mole Fraction=
Moles solute/total number of moles
Mass %=
Mass solute/total mass x 100
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15. Determine the mole fraction of KCl in 3000 grams of
aqueous solution containing 37.3 grams of Potassium Chloride KCl.
Convert grams KCl to moles KCl using the molecular weight of KCl
2. Determine the grams of pure solvent water from the given grams of solution and solute
Total grams = 3000 grams = Mass of solute + Mass of water
Mass of pure solvent = ( ) gram
= gram
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16. Determine the mole fraction of KCl in 3000 grams of
aqueous solution containing 37.3 grams of Potassium Chloride KCl.
3. Convert grams of solvent H2O to mols
4. Apply the definition for mole fraction mole fraction = moles of KCl / Total mols of KCl and water =
0.5 / ( ) = 0.5 / =
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17. Mole Fraction

18. Exercise #5
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.)
0.0145

19. Mass Percent

20. Exercise #6
What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?
6.6%

21. Assuming the density of water to be 1 g/mL we approximate the density of a dilute aqueous solution to be 1 g/mL
 1 ppm = 1 μg/mL = 1 mg/L
 1 ppb = 1 ng/mL = 1 μg/L
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22. Molarity Concentration is usually expressed in terms of molarity:
Moles of solute/liters of solution (M)
Moles of solute = molarity x volume of solution
Moles = M x V

23. Molarity and concentration
M = moles solute/liter of solution
Dilution
M1V1 = M2V2

24. Example What is molarity of 50 ml solution containing 2.355 g H2SO4?
Molar mass H2SO4 = 98.1 g/mol
Moles H2SO4 = mol (2.355 g/98.1 g/mol)
Volume of solution = 50 mL/1000 mL/L = .050 L
Concentration = moles/volume
= mol/.050 L = M

25. Solution stoichiometry
How much volume of one solution to react with another solution
Given volume of A with molarity MA
Determine moles A
Determine moles B
Find target volume of B with molarity MB