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

• χsolvent = #mol of solvent

#mol of solution

• Unit-less

Sum willbe one

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 andpour into the 250-mL volumetric flask. Fillthe 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