Solutions Molarity, Molality, Dilutions, Percent Solutions, & Mole Fractions.
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Transcript of Solutions Molarity, Molality, Dilutions, Percent Solutions, & Mole Fractions.
![Page 1: Solutions Molarity, Molality, Dilutions, Percent Solutions, & Mole Fractions.](https://reader035.fdocuments.net/reader035/viewer/2022062308/56649d8e5503460f94a77d8a/html5/thumbnails/1.jpg)
SolutionsMolarity, Molality, Dilutions, Percent Solutions, & Mole
Fractions
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Concentration
Concentration is a value that represents the amount of solute dissolved in a solvent.
Concentrated solutions have a large amount of solute relative to the solvent
Dilute solutions have a small amount of solute relative to the solvent
Concentration can be measured in Molarity, molality, %solution, ppm (parts per million)
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How it is made…
2 Molar Solution of NaClrequires 2 moles of NaCl to be dissolved to
make 1L of solution2 moles of salt (or 117 grams) are dissolved in 1L of
water to make a 2M NaCl(aq) solution
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Molarity
Molarity is the moles of solute per liter of solution.Note the volume is the total solution volume,
not the volume of solvent alone.
M = moles = mass/molar mass
Liters Liters
This is the most common way chemists measure concentration
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Practice Problem #1
What is the molarity of HCl if 28g HCl is dissolved in 500.mL of solution?
mass=28.g molar mass of HCl= 36.46g/mol
V= 500mL=0.500L (divide by 1000)
M= moles = (mass/molar mass)Liter Liter
= (28/36.46)moles
0.500L
= 1.5 M
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Practice Problem #2
How many grams of sodium nitrate (NaNO3 ) are needed to make 2 liters of a 0.100M solution?
M=0.100M m=? mm NaNO3 = 85g/mol
0.100M = (m/85)mols 2L
(0.100M)(2L)(85)= m
m=17g NaNO3
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Solving using Dimensional Analysis
Molarity can be used as a conversion factor.
2.0 L 0.100moles 85 grams = 17g
1 Liter 1 moles
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Example 3What is the molarity of a solution with 10.0
grams of AgNO3 is dissolved in 500.mL of solution.
Molarity = Mass / Molar Mass Liters of Solution
M = 10.0g / 169.88 g/mol
0.500 Liters
Must convert to Liters!!
500mL/ 1000
= 0.500 Liters
M = 0.118 Molar
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Example 4
How many grams of KNO3 should be used to prepare 2.00 Liters of a 0.500M solution?Molarity = Mass / Molar Mass Liters of Solution
0.500 = mass / 101.11 2.0
(0.500)(2.0) = Mass 101.11
Mass = (0.500)(2.0)(101.11)Mass = 101.11 grams
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Example 5
To what volume should 5.0 grams of KCl be diluted to in order to prepare a 0.25M solution?
Molarity = Mass / Molar Mass Liters of Solution
0.25 = 5.0 / 74.55 Volume
V(0.25) = 5.0/74.55
V = 0.0671 0.25
V= 0.268 Liters = 268 mL
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Molality (m)Molality is the concentration of a solution expressed
in moles of solute per kilogram of solvent.
moles of solute (mol)Molality (m) = -----------------------------------
kilogram of solvent (kg)
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Molality Sample Problem
A solution was prepared by dissolving 17.1 g of glucose, C6H12O6, in 275 g of water. What is the molality (m) of this solution?
0.345 mol/kg or 0.345 m
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More molality examples
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Preparing Solutions from Other SolutionsMany times solutions are prepared by diluting
concentrated solutions. These are called stock solutions.
Dilution = process of adding water to a stock solution to achieve the molarity desired for a particular solution.
Adding water increases the volume of the solution and this causes the concentration to decease.
Dilution Equation:M1V1 = M2V2
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Dilutions and Molarity
Use this formula to make a more dilute solution from a concentrated solution
Molarity1Volume1 = Molarity2Volume2
(Concentrated) (Dilute)
(Moles before) = (Moles after!)
M1V1 = M2V2
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Sample Problem #1
How many milliliters of 1.0 Molar HCl are required to make 100 mL of 0.025 M HCl?
M1V1 = M2V2
M1 = 1.0M V1 = ? M2 = 0.025M V2 = 100mL
(1.0M)V1 = (0.025M)(100mL)
V1 = (0.025M)(100mL) (1.0M)
= 2.5 mL of 1.0M HCl required, then dilute to 100mL
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Sample Problem #2How much water should you add to the
volume of 1.0M HCl you calculated above to make the solution?
We need 2.5 mL of concentrated 1.0M HCl. Then we dilute to a final total volume of 100mL.
Therefore, 100mL – 2.5mL = 97.5 mL of water should be added.
**Water added = V2 – V1
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More dilution example problems
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Percent Solutions
Solutions can also be represented as percent of solute in a specific mass or volume of solvent.
For a solid dissolved in water, you use percent by mass.
% by mass = mass solute x 100 mass of solution
*Mass of solution = solute mass + solvent mass
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For a liquid mixed with another liquid
% by volume = volume solute x 100
Total volume of solution
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Example 1What percent solution do you have if you
dissolve 80 grams of NaCl in 1 liter of water?
* 1 Liter = 1000mL = 1000grams for water
% mass = 80 grams NaCl X 100
(80g + 1000g H2O)
= 7.4 %
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Example #2
What percent solution will you have if you mix 40mL of ethanol with 200 mL of water?
%volume = 40mLs x 100
(200mLs + 40mLs)
= 16.7%
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What if the previous problem was worded as follows?What percent solution will you have if you
dilute 40mL of ethanol to a final volume of 200mL?
% volume = 40mL x 100
200mL*
= 20%
*Denominator is always total volume. You must pay attention the wording in the problem!
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More Percent Solution Problems
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Mole Fraction (X)
Number of moles of a compound divided by the total number of moles of all species in the solution.
Adding all mole fractions should get you a value of 1
Guess what the units are?
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Example
What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molar mass of glycerol =92 g/mol)
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Answer:
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Example 2
What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass? (molar mass of NaOH is 40 g/mol)
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Answer:
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More Mole Fraction Examples