The concentration of a solution is referred to as the ratio of the solute amount to the amount of the solution (or a solvent in some specific cases).
The below is the table summarizing all types of concentrations formulas used in this chapter
Name of the Concentration | Formula | Unit of solute | Unit of solution (solvent) | Overall unit of the formula |
(Mass/volume)% | ={[mass of the solute]/[volume of the solution]} x100% | Grams (g) | Milliliters (mL) | [g/mL]% |
(Mass/mass)% or mass% | =[mass of the solute]/[mass of the solution]x100%Note that:Mass of solution = mass of solute + mass of solvent | Grams (g) | Grams (g) | [g/g]% |
(Volume/volume)% or volume% | =[volume of the solute]/[volume of the solution]x100%Note that:Volume of the solution = volume in solute + volume of the solvent | Milliliters (mL) | Milliliters (mL) | [mL/mL]% |
(Molarity) | = [moles of the solute]/[liter of solution] | Moles (mol) | Liter (L) | [mol/L]Also can be expressed as M or molar |
(Molality) | = [moles of the solute]/[Kilogram of solvent | Moles (mol) | Kilogram (kg) | [mol/kg]Also can be expressed as m or molal |
Equivalent molarity of the electrolytes | = [milliequivalents/Liter of solution] | Milliequivalents (mEq) | Liter (L) | [mEq/L] |
Dilution | C1xV1 = C2xV2 or sometimes is given in:M1xV1=M2xV2Where C1 and M1 are the concentration and the molarity of the stock solution respectively. C2 and M2are the concentration and molarity of the diluted solution respectivelyV1 is the volume of the stock solution and V2 is the volume of diluted solution | Moles (mol) | Liter (L) | [mol/L]Also can be expressed as M or molar |
Let us now go over some workout problems in some details covering all the formulas.
Q: What is the mass/volume percent by of a solution formed by mixing 25.0 grams of sodium chloride NaCl with 455 mL of water?
Solution:
(Mass/volume)% = {[mass of the solute]/[volume of the solution]} x100%
= {[25.0 g NaCl] / [455 mL solution]} x100% = 5.494505494505495% (answer from calculator and it should be rounded off to the correct significant figures which is 3.
Final answer after the significant figures rule consideration is 5.50% (g/mL %)
Q: Calculate amount of grams of solute potassium nitrate aqueous solution which is 38.5% (mass/mL%) if the solution made with 1.00 Liter of water.
Solution:
Let us set up the formula and replacing it with the given data:
(Mass/volume)% = {[mass of the solute]/[volume of the solution]} x100%
First the volume should be converted from Liter into milli Liter (mL)
[38.5 g/mL]% = { [X] / [1000 mL] } x 100%
Cross x Multiply
[38.5 g/mL] / [100] = [X] / [1000 mL]
X = 385 g potassium nitrate KNO3
Verifying the answer:
(Mass/volume)% = {[mass of the solute]/[volume of the solution]} x100%
= [385 g / 1000 mL] x100% = (38.5 g/mL)%
Conclusion the answer is correct.
Q: A solution is formed by adding 48 g of Lithium sulfate to 250 g of water. What is the percent by mass of Lithium sulfate?
(Mass/mass)% or mass% = [mass of the solute]/[mass of the solution]x100%
Note that:
Mass of solution = mass of solute + mass of solvent
(Mass/mass)% = {[48 g of Lithium sulfate] /[48 g of Lithium sulfate + 250 g water]} x 100%
= 16.10738255033557% from calculator and it should be rounded off to the correct significant figures
Final answer = 16% [g/g]%
Q: What is the mass of solvent (water) in grams for a solution made of 10.5% mass percent of each component in the mixture formed by adding 2.00 g of Aluminum nitrate in water?
(Mass/mass)% or mass% =[mass of the solute]/[mass of the solution]x100%
Note that:
Mass of solution = mass of solute + mass of solvent
Mass of the whole solution is not known and it will be referred to as X
[10.5 / 100] = [2.00 g Aluminum nitrate] / [ X ]
Cross / Multiply:
[0.105] * [ X ] = 2.00
X = [2.00 /0.105] = 19.04761904761905 = 19.0
X = amount of water in grams = 19.0 – 2.00 = 17.0 grams water
Verifying the answer:
[Mass/mass]% = [2.00 /(2.00 + 17.0)] x 100% = 10.52631578947368% = 10.5%
The answer is correct.
Q: What is the percent by volume of a solution formed by added 25 L of ethanol to 155 L of water?
(Volume/volume)% or volume% = [volume of the solute]/[volume of the solution]x100%
Note that:
Volume of the solution = volume in solute + volume of the solvent
(Volume/volume)% = [ 25 L ] /[ 25L + 155 L ] x 100 % = [ 25 L ] / [ 180 L ] = 13.88888888888 %. The answer is from calculator and it should be rounded off to the correct significant figures
Final answer is 14 % (two significant figures).
Q: A 5.50% volume percent solution is made by adding certain amount of volume of a solute acetone to 255 mL water as solvent. Calculate the amount of the acetone in mL added.
(Volume/volume)% or volume% = [volume of the solute]/[volume of the solution]x100%
Note that:
Volume of the solution = volume in solute + volume of the solvent
Amount of the solute is unknown and it will be referred to as X
5.50% = [ X ] / [ X + 255 mL ]
0.0550 = [ X ] / [ X + 255 mL ]
0.0550 [ X + 255 mL ] = [ X ]
0.0550 X + (0.0550*255) = [ X ]
Solving for X
0.0550*255 = X – 0.0550 X = 0.945 X
X = 14.8 mL (3 significant figures)
Verifying the answer:
Volume/volume)% = { [ 14.8 mL ] / [ 14.8 mL + 255 mL] } x 100% = 5.485544848035582 % rounded to 3 significant figures = 5.50 %. The answer is correct.
Molarity = [moles of the solute]/[liter of solution]
Q: Calculate the molarity of solution made by dissolving 35.0 grams of sodium chloride solid in 97890. mL distilled water.
Molarity = [moles of the solute]/[liter of solution]
First one has to calculate the number of moles of sodium chloride with given mass of 35.0 grams. Therefore, one has to use the molar mass of sodium chloride in calculating number of moles with given number of grams.
Molar mass of sodium chloride = NaCl using the atomic masses of Na and Cl, the molar mass is:
23.0 g/mole + 35.5 g/mol = 58.5 g/mol
Number of moles of sodium chloride = (35.0 g) / (58.5 g/mol) = 0.598 moles NaCl
Second one has to convert the mL into Liters by dividing by 1000 =(97890. mL) / (1L / 1000 mL) = 97.890 L
Molarity = [0.598 moles NaCl] / [97.890 L} = 0.006108897742363877 mol/L. This the calculator’s answer and it will be rounded off to the correct significant figures ( 3 significant figures).
Molarity = 0.00611 mol/L or 0.00611 M or 0.00611 molar.
Q: A solution of a molarity of 2.675 M is prepared by dissolving 8.50 gram of magnesium nitrate in certain amount of distilled water as a solvent. Calculate the amount of the distilled water.
First one has to calculate the number of moles of magnesium nitrate with given mass of 8.50 grams.
Therefore, one has to use the molar mass of magnesium nitrate in calculating number of moles with given number of grams.
Molar mass of magnesium nitrate = Mg(NO3)2 using the atomic masses of Mg and 2 (NO3), the molar mass is: 24.3 g/mol + 2x (14.0 g/mol) + 2 (3 x 16.0 g/mol) =148.3 g/mol
Number of moles of magnesium nitrate = [8.50 g Mg(NO3)2] / [148.3 g Mg(NO3)2 / mol Mg(NO3)2]
Number of moles of magnesium nitrate = 0.0573 moles
2.675 M = 2.675 mol /L = [0.0573 mol] / [X]
X = amount of the solution or solvent in Liters
Solving for X, one can obtain:
2.675 X = 0.0573
X = 0.0573 / 2.675 = 0.02142056074766355 Liters. The correct rounded off answer has to have three significant figures. The final correct answer is 0.0214 Liters
Verifying the answer: = [0.0573 mol] / [0.0214 Liters] = 2.6775700934579439 = 2.678 mol/L
The answer is correct. Note that the verified answer is not exactly the given molarity of 2.675 mol/L. This is because of the in between rounding off and rounding off the atomic masses.
Q:10.00 x 1026 molecules of acetone dissolved to make 3500. mL of water of solvent. Calculate the molarity of acetone solution.
First one has to calculate the number of moles of acetone with given number of molecules.
Number of moles of acetone = [10.00 x 1026 molecules of acetone] x [1 mole acetone/6.022 x 1023 molecules of acetone]
One has to use the Avogadro’s number of 6.022 x 1023 to calculate number of moles:
1 mole of acetone = 6.022 x 1023 molecules of acetone
Moles of acetone = 1660.577881102624 moles
The volume of solution is 3500. mL = 3.500 L (volume is converted into Liters)
Molarity = [1660.577881102624 moles] / [3.500 L] = 474.4508231721783 mol/L = 474.5 M or 474.5 mol/L or 474.5 molar.
The answer is rounded to the correct 4 significant figures.
Q: What is the molality of a solution made of 9.384 grams of aluminum nitrate has been dissolved in 5500.0 g water
First the number of moles of aluminum nitrate is calculated using the molar mass of aluminum nitrate.
Molar mass of aluminum nitrate= Al(NO3)3 = Al + 3N + 9O = 27.0 g/mol + (3 x 14.0) g/mol + (9 x 16.0 g/mol) = 213.0 g/mol
Number of moles = [9.384 grams of aluminum nitrate] / [mol aluminum nitrate /(= 213.0 g/mol)] = 0.04405633802816902 moles aluminum nitrate.
Second to convert 5500.0 g water into kilo grams water by diving by 1000 = 5.500 kg water
Molality = [0.04405633802816902 moles aluminum nitrate] / [5.500 kg water] = 0.008010243277848913 mol/kg. The answer should be rounded off to the correct significant figures of 4.
The final correct answer is 0.008010 mol/kg or m or molal.
Q: What mass of water is required to dissolve 45.0 grams calcium chloride to prepare a 2.50 m solution?
First the number of moles of calcium chloride is calculated using the molar mass of calcium chloride.
Molar mass of calcium chloride CaCl2 = Ca + 2 Cl = 40.0 g/mol + 2x 35.5 g/mol = 111.0 g/mol
Number of moles CaCl2 = [45.0 g calcium chloride] / [111.0 g/mol] = 0.4054054054054054 moles = 0.405 moles CaCl2
Second the molality formula is used:
2.50 m = [2.50 CaCl2 mol / kg water] = [0.405 moles CaCl2] / X
Where X is the amount of water
Solving for X yields:
X = 0.405 / 2.50 = 0.162 kg water
Verifying:
Molality = [0.405 moles CaCl2] / [0.162 kg water] = 2.50 m
Milliequivalents (mEq) calculations are covered in greater details in the link below:
Milliequivalents calculations cover the three topics:
Each electrolyte such as Na+ is measured in equivalent (Eq).For example 1 mole of an ionic compound of NaCl dissolved in water will produce 1 mole of Na+ 1 mole of and Cl–. Each Na+ and Cl– which equal to 1 equivalent each.
The table below illustrates the relationship between the electrolytes charge and number of the equivalents.
ion | Ion charge | Ion Oxidation number | Number of Equivalents per 1 mole |
Li+, Na+, K+NH4+ | 1+ | +1 | 1 Eq |
Mg2+, Ca2+, Ba2+ | 2+ | +2 | 2 Eq |
Al3+, Cr3+, Fe3+ | 3+ | +3 | 3 Eq |
Br–, I–, F– | 1- | -1 | 1 Eq |
CO32-, SO42- | 2- | -2 | 2 Eq |
PO43- | 3- | +3 | 3 Eq |
What is the concentration of a solution contains 8.50 mEq/L of NaCl?
First the molar mass of NaCl should be calculated. NaCl molar mass is calculated in the above examples and it equals 58.5 g/mol
Second use the general formula for the calculation of the milligram/L from mEq/L:
Mg/L = [(mEq/L) x (atomic mass or molar mass)] / [number of the equivalent)
Mg/L = [(8.50 mEq/L) x (58.5 g /mol)] / [1 Eq/ 1 mol] = 497.25 mg / L
Final answer is 497 mg / L (3 significant figures).
How many mEq of NaCl are in 10.5 g of NaCl?
First the molar mass of NaCl should be calculated. NaCl molar mass is calculated in the above examples and it equals 58.5 g/mol
Second use the setup that relates mEq to mass of electrolytes using molar mass:
mEq = {[10.5 g of NaCl] / [58.5 g NaCl/mol]} * [Eq /1 mol] * [1000 mEq/ 1Eq] = 179 mEq
Convert the expression 15.0 mg% of Ba2+ to mEq/L
Ba2+ atomic mass = 137.3 g/mol
Ba has 2+ ion and hence 2 Eq
Equivalent mass of Ba2+ = [137.3 g/mol] / [2 Eq/mol] = 68.65 g/Eq = 68.7 g per Eq
1mEq Ba2+ = [68.7 g] / [1 g/1000 mg] = 0.0687 mg
15.0 mg% = [15.0 mg Ba2+] / [100 mL solvent] = [15.0 mg/ 100 mL]*[1000 mL/1L] = 150. mg/L
0.0687 mg/ 1 mEq = 150. mg/ X mEq
Solving for X = 150. / 0.0687 = 2183.406113537118 mEq/L rounded off to 2180 mEq/L (3 significant figures) or expressed in scientific notation 2.18 X 103 mEq/L
C1xV1 = C2xV2 or sometimes is given in:
M1xV1=M2xV2
Where C1 and M1 are the concentration and the molarity of the stock solution respectively. C2 and M2are the concentration and molarity of the diluted solution respectively
V1 is the volume of the stock solution and V2 is the volume of diluted solution
The stock solution is considered as the most concentrated solution
Examples:
If 75.5 mL of water are added to 875.0 mL of a 0.125 M HNO3 solution, what will be the new molarity of the diluted solution?
C1xV1 = C2xV2
Total volume = V2 = 75.5 + 875.0 = 950.5 mL
0.125 M x 875.0 mL = C2 x 950.5 mL
C2 = [0.125 M x 875.0 mL] / [950.5 mL] = 0.115 M
If 500.0 mL are added to 0.150 M HCl solution until the final volume is 650.0 mL, what will be the molarity of the diluted solution?
C1xV1 = C2xV2
Total volume = V2 = 650.0 mL
0.150 M x 500.0 mL = C2 x 650.0 mL
C2 = [0. 0.150 M x 500.0 mL] / [650.0 mL] = 0.115 M
C1xV1 = C2xV2
Total volume = V2 = X + 985 mL
X = volume of water
6.55 M x 985 mL = 1.00 M x V2
V2 = [6.55 M x 985 mL] / [1.00 M] = 6451.75 mL
X + 985 mL = 6451.75 mL
X = 5466.75 mL = 5470 mL rounded off to 3 significant figures
C1xV1 = C2xV2
Total volume = V2 = 450.5/2 mL = 225.3 mL
450.5 mL x 2.660 M KNO3 = C2 x 225.3 mL
C2 = [450.5 x 2.660] / [225.3] = 5.319 M [Boiling the stock solution had led to higher concentration of the new boiled solution]
After going over couple examples of the concentrations, let us look at some available simulation activities for the concentrations of the solution.
A Phet molarity simulation activity is given below:
In this simulation, the students will use the information below:
Solution volume is the combined volume of solute and water.
By design, not all solutions will reach saturation. The number of moles that can be added is limited to the range of 0.2-1.0 moles so that students can explore some solutions for the full concentration range (0-5 M).
Drink mix is assume to have the same solubility as sucrose.
Solubility of each solution listed was calculated at 250 C, except for AuCl3 and Drink mix, which were based on data taken at 200 C.
Activity:
Determine the qualitative relationships between molarity, moles, and liters before completing quantitative problems or data collection.
The simulation demonstrates saturation but does not explain why different solutes have different solubilities. The Drink Mix example provides a real-world link to the concept of concentration to help the students make connections to the chemical examples. Determine the saturation molarity in mol/L for all salts (solutes) in the simulation.
Several You Tube videos discussing the concentration of solution:
(Mass) %https://www.youtube.com/embed/lywmGCfIUIA?feature=oembed
(Mass/volume) %:https://www.youtube.com/embed/dl6G_p6gsJc?feature=oembed
(Volume/volume) %:https://www.youtube.com/embed/RCbhk3yyM88?feature=oembed
Molarity:https://www.youtube.com/embed/SXf9rDnVFao?feature=oembed
Molality:
https://www.youtube.com/watch?v=uj1u7Nx9JUc
Dilution:
Dilution:
the component of a solution present in a greatest quantity
the component of a solution present in a least quantity
A homogenous mixture of two or more substance in which wach substance retains its chemical identity
the amount of solution in a specified amount of solution
Percent by mass%(m/m)
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%(v/v)
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%(m/v)
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Molarity
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