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Chapter 11 1
3. Units of Concentration
molestotal
Amolesunitless
%100masstotal
Amass
Definition
same
same
Chapter 11 2
Advantages/disadvantages of each (read only; no lecture)
Chapter 11 3
Example problems1. Determine the molarity of a 0.258 m solution of glucose given that the solution’s density is 1.0173 g ml-1 and that the molar mass of glucose is 180.2 g.
•first, write down definitions: molarity is moles solute per liter solution; molality is moles solute per kg solvent; density is mass solution per volume solution
•assume some amount of solution to start: in this case 0.258 moles glucose and 1 kg water
we want: moles solute/liter solution
so we have 0.258 moles solute
Chapter 11 4
•get the total volume of the solution - glucose plus water
•solve for the molarity
nsol'1046.49water1000glucose46.49
glucose49.462.180
258.0
ggg
gmol
gmoles
mlg
mlnsolg 70.1028
0173.1
1'49.1046
mass sol’n
volume sol’n
ML
moles251.0
02870.1
258.0
Chapter 11 5
2. Determine the molality of a 0.500 M solution of acetic acid (molar mass 60.02 g) with a density of 1.0042 g ml-1.
•Again, write down the units and their definitions.
•Start by assuming 1 L of solution and thus 0.500 moles acetic acid
•determine the mass of the solution from the density times the volume
•determine the mass of water (in kg) by subtraction (subtract the mass of the acetic acid from the mass of the solution)
•solve for the molality
Chapter 11 6
3. Describe how 1.50 L of a 12.0% KBr solution is made if the solution density is 1.10 g ml-1. In other words, determine the mass of KBr and the mass of water to be mixed together.
•Write out the units and their definitions
•1.50 L or 1500 ml of solution are required; determine the total mass of solution required (HOW?).
•Of the total mass of solution, how much must be KBr?
•Determine the mass of water necessary (HOW?).
Chapter 11 7
gmlgml 165010.11500
KBr198165012.0 g
water1452KBr198nsol'1650 ggg
total mass solution
mass KBr
mass water
dissolve 198 g KBr in 1452 g water