solutions homogeneous mixture solute solvent. solution formation rate factors affecting it…...
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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.