II III I II. Solution Concentration (p. 480 – 488) Ch. 14 – Mixtures & Solutions.
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Transcript of II III I II. Solution Concentration (p. 480 – 488) Ch. 14 – Mixtures & Solutions.
![Page 1: II III I II. Solution Concentration (p. 480 – 488) Ch. 14 – Mixtures & Solutions.](https://reader036.fdocuments.in/reader036/viewer/2022062404/5514ed4e550346b0338b5e0b/html5/thumbnails/1.jpg)
II
III
I II. Solution Concentration
(p. 480 – 488)
Ch. 14 – Mixtures & SolutionsCh. 14 – Mixtures & Solutions
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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
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A. MolarityA. Molarity
Concentration of a solution
solution of liters
solute of moles(M)Molarity
total combined volume
substance being dissolved
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A. MolarityA. Molarity
2M HCl
L
molM
nsol' L 1
HCl mol 2HCl 2M
What does this mean?
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Molar Mass(g/mol)
6.02 1023
particles/mol
MASSIN
GRAMSMOLES
NUMBEROF
PARTICLES
Molar Volume (22.4 L/mol)
LITERSOF GASAT STP
LITERSOF
SOLUTION
Molarity(mol/L)
B. Molarity CalculationsB. Molarity Calculations
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B. Molarity CalculationsB. 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
58.44 g NaCl
1 mol NaCl
L 1
mol0.25 0.25M
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B. Molarity CalculationsB. 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= 0.95 M
NaF
L
molM
.25 L sol’n
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2211 VMVM
C. DilutionC. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.
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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
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D. MolalityD. Molality
solvent ofkg
solute of moles(m)molality
mass of solvent only
1 kg water = 1 L water
kg 1
mol0.25 0.25m
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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
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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
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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
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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!
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Solution Preparation Mini-LabSolution Preparation Mini-Lab
Turn in one paper per team. Complete the following steps:
A) Show the necessary calculations.
B) Write out directions for preparing the solution.
C) Prepare the solution. For each of the following solutions:
1) 100.0 mL of 0.50M NaCl
2) 0.25m NaCl in 100.0 mL of water
3) 100.0 mL of 3.0M HCl from 12.0M concentrate. (don’t actually prepare this one!)