Solutions
• Homogeneous mixture
• Solute
• Solvent
Solution Formation Rate
• Factors affecting it…
– Temperature—think about
dissolving sugar in tea
– Agitation—again think about the tea
– Particle size—granulated sugar vs.
powdered sugar?
Henry’s Law
• The solubility (C) of a gas in a
liquid is directly proportional to the
pressure above the liquid
• C1 = C2
P1 P2
• Think about carbonated beverages
Henry’s Law Problem
• At 20°C and 1.00atm, the
solubility of oxygen gas in water is
0.0448g/L. What will the
solubility be if the pressure is
increased to 1.75atm?
Answer: 0.0784g/L
Concentration
• Qualitative descriptions do not
give amounts of solute in solution.
– Concentrated—lots of solute
– Dilute—not much solute
• Note: no quantities are provided
Concentration
• A few types…
– Molarity
– Mass percent
– Mass/volume percent
– Volume/volume percent
– Mole fraction
– Molality
Molarity
• Represented by M—always
uppercase
• M = #moles of solute
#liters of solution
• Units will be mol/L, or you may
write it as M
• A 2M solution is described as “two
molar”
Mass Percent
• Represented as m/m%
• m/m% = mass of solute x 100
mass of solution
• Mass units must be the same for
the solute and the solution
• Unit-less
Mass/volume Percent
• Represented as m/v%
• m/v% = #g of solute x 100
#mL of solution
• Units are specific
• g/mL
Volume/volume Percent
• Represented by (v/v)%
• (v/v)% = volume of solute x 100
volume of solution
• Volume units must be the same for
the solute and the solution
• Unit-less
Mole Fraction
• Represented by Greek letter, chi, χ
• χsolute = #mol of solute
#mol of solution
Sum will
be one
• χsolvent = #mol of solvent
#mol of solution
• Unit-less
Molality
• Represented by m—always
lowercase
• m = #moles of solute
#kg of solvent
• Units will be mol/kg, or you may
write it as m
• A 2m solution is described as “two
molal”
Dilutions
• I don’t usually stock every
concentration of every acid that I
need…so I start with the most
concentrated form and make
whatever molarity (the most
common measurement of
concentration) I need.
• (M1)(V1) = (M2)(V2)
Dilutions
• The rule of thumb is if you’re
making an acid dilution in a
beaker, always add the acid to
water…so if something splashes, it
will be the water!
Practice #1
• How many grams of sodium
hydroxide are required to make
250mL of a 0.25M solution?
Answer: 2.50g NaOH
Practice #2
• How many grams of water are
present in a 5.00%(m/m) solution
containing 0.875g of calcium
acetate?
Answer: 16.625g H2O
Practice #3
• How many grams of acetic acid are
present in 4.50L of a 5.00%(m/v)
solution?
Answer: 225g HC2H3O2
Practice #4
• A 95% (v/v) solution of ethanol in
double-distilled water is used to
clean surfaces in a laboratory
setting. If you have 500mL of the
solution, how many of those mL
are water?
Answer: 25mL H2O
Practice #5
• 10.0g of calcium acetate are added
to 100g of water. What are the
mole fractions of both the solute
and the solvent?
Answer: χcalcium acetate = 0.0114
χwater = 0.989
Practice #6
• 10.0g of calcium acetate is added
to 100g of water. What is the
molality of this solution?
Answer: 0.632mol/kg or m
Practice #7
• If I need to make 250mL of 6M
sulfuric acid, and all I have in my
cabinet is full strength, 18M
sulfuric acid, describe how I make
the dilution.
Answer: measure 83.3mL of 18M H2SO4 and
pour into the 250-mL volumetric flask. Fill
the flask to the etched line with water, cork it,
and agitate it.
A Doozie Problem
• A solution is prepared by mixing 1.00g
of ethanol, CH3CH2OH, with 100.0g of
water to give a final volume of
101.0mL of solution. Calculate the
molarity, mass percent, mass/volume
percent, molality, mole fraction of the
ethanol, and mole fraction of the water.
A Doozie’s Answers…
• molarity = 0.215M or 0.215mol/L
• mass percent = 0.990%
• mass/volume percent = 0.990(g/mL)%
• molality = 0.217m or 0.217mol/kg
• mole fractionethanol = 0.00389
• mole fractionwater = 0.996