Post on 16-Jan-2016
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II
III
I II. Solution Concentration
(p. 480 – 488)
II. Solution Concentration
(p. 480 – 488)
Ch. 14 – Mixtures & SolutionsCh. 14 – Mixtures & Solutions
A. ConcentrationA. Concentration
The amount of solute in a solution
Describing Concentration
• % by mass - medicated creams
• % by volume - rubbing alcohol
• ppm, ppb - water contaminants
• molarity - used by chemists
• molality - used by chemists
B. Percent by MassB. Percent by Mass
Percent Composition by Mass is the mass of the solute divided by the mass of the solution (mass of the solute plus mass of the solvent), multiplied by 100.
Mass of SoluteMass of Solution
x100
B. Percent by MassB. Percent by Mass
How many moles of solute are contained in 343 grams of a 23% aqueous solution of MgCr2O7?
343g of solution 23%
100%
1 mol
240.3 g MgCr2O7
=0.329 mol of MgCr2O7
C. Percent by VolumeC. Percent by Volume
Percent Composition by Volume is the volume of the solute divided by the volume of the solution (volume of the solute plus volume of the solvent), multiplied by 100.
Volume of SoluteVolume of Solution
x100
B. Percent by VolumeB. Percent by Volume
Determine the percent by volume of toluene (C6H5CH3) in a solution made by mixing 40.0 mL of toluene with 75.0 mL of benzene (C6H6).
40.0 mL toluene + 75.0 mL benzene= 115 mL total
solution(40.0 mL toluene / 115 mL solution) 100 = 34.8%
toluene
C. MolarityC. Molarity
Concentration of a solution
solution of liters
solute of moles(M)Molarity
total combined volume
substance being dissolved
C. MolarityC. Molarity
2M HCl
L
molM
nsol' L 1
HCl mol 2HCl 2M
What does this mean?
C. Molarity CalculationsC. Molarity Calculations
How many grams of NaCl are required to make 0.500L of 0.25M NaCl?
0.500 L sol’n 0.25 mol NaCl
1 L sol’n
= 7.3 g NaCl
=.125 mol NaCl
58.44 g NaCl
1 mol NaCl
.125 mol NaCl
C. Molarity CalculationsC. Molarity Calculations
Find the molarity of a 250 mL solution containing 10.0 g of NaF.
10.0 g NaF 1 mol NaF
41.99 g NaF
L
molM .283 NaF
.25 L sol’n= 0.95 M
NaF
=.283 mol NaF
D. MolalityD. Molality
solvent ofkg
solute of moles(m)molality
mass of solvent only
1 kg water = 1 L waterkg 1
mol0.25 0.25m
D. MolalityD. Molality
Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.
75 g MgCl2 1 mol MgCl2
95.21 g MgCl2
= 3.2m MgCl2
0.25 kg water
kg
molm
D. MolalityD. Molality
How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water?
0.500 kg water 1.54 mol NaCl
1 kg water
= 45.0 g NaCl
58.44 g NaCl
1 mol NaCl
kg 1
mol1.54 1.54m
E. Mole FractionE. Mole Fraction
The number of moles of a component of a solution divided by the total number of moles of all components.
Moles A
Total Moles= Mole Fraction, Χ
2211 VMVM
C. DilutionC. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.
C. DilutionC. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
E. Preparing SolutionsE. Preparing Solutions
500 mL of 1.54M NaCl
500 mLwater
45.0 gNaCl
• mass 45.0 g of NaCl• add water until total
volume is 500 mL• mass 45.0 g of NaCl• add 0.500 kg of water
500 mLmark
500 mLvolumetric
flask
1.54m NaCl in 0.500 kg of water
E. Preparing SolutionsE. Preparing Solutions
250 mL of 6.0M HNO3 by dilution
• measure 95 mL of 15.8M HNO3
95 mL of15.8M HNO3
water for
safety
250 mL mark
• combine with water until total volume is 250 mL
• Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid!