In this section, we"re largely working with aqueous (water-as-solvent) solutions, however the same actions of concentration can apply to any type of other solvent. Once the solven is not given, or if we acknowledge the solutes as water-soluble (solutes most generally dissoved in water), we assume that the equipment is aqueous.

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Strength the a solution

We require a method to represent the family member amount of solute liquified in a solvent.

It provides sense that some remedies are more powerful than others: including a tiny little bit of mountain to a gallon of water, for example, could not even be detectable. But include a lot the acid and you have actually a dangerous systems that requirements to be tackled carefully.

We need a way to stand for the loved one strength of a solution. We"ll call it concentration.

There space several methods of concentration measurement, each provided in different kinds the situations.


If we dissolved just a pair of crystals of table salt (NaCl) in water, we can not even be able to taste it. Yet if we dissolved a bunch the NaCl in the same amount of water (and you could be surprised exactly how much will dissolve), it would certainly taste an extremely salty.

On a relative scale, one solution is dilute and also the other is concentrated. We"d favor to have the ability to put this on some type of numerical scale so the we could say just precisely how dilute or concentrated a systems is.

Remember that the solute (usually a solid) is what"s being dissolved in a solvent (usually a liquid)

All of the concentration measurement techniques covered listed below consist of some measure that the amount of solute (in grams, moles or atoms/molecules) divided by the quantity of solvent (in units of mass, volume, mole or number of atoms/molecules).

1. Molarity (M)

Molarity, abbreviated (M) is most likely the most frequently used measure of solution concentration. Molarity is the variety of moles of solvent divided by the variety of liters that solution. We refer to a solution, for example, as a "1.5 molar solution" or "1.5 M".

Note that we divide by the total number of liters that solution, including the solute, not the variety of liters of solvent to i m sorry the solute was added — it"s vital (if often small) distinction.

helpful tip:

This figure ( → ) shows just how to make an X-molar (X M) solution, wherein X is the desired molar concentration. We have actually a target volume (1 liter in this case, yet it can be anything). The solute is liquified in a smaller volume that solvent, then the complete volume is changed to the final desired amount.



Molarity (M) is the number of moles of solute separated by the total volume of the systems in liters. A 1 M solution has 1 mole that solute because that every 1 l of solution.

Example 1

Calculate the molarity that a NaCl solution developed by dissolve 62 g that NaCl in water and also adjusting the full volume come 0.50 liters.

Solution: begin by calculating the variety of moles that solute


Then division it by the total number of liters of solution, 0.5 together in this case:


Example 2

How numerous grams that (NH4)2SO4 (ammonium sulfate) should be added to water to make 200 ml the a 1.6 M solution?

In this problem we recognize the concentration, we just need to understand the lot of solute required to attain it. Fist uncover that number of moles:


Then convert it to grams.


Note the in making this solution, we"d want to dissolve the solute in less than 200 ml the water, then carry the complete volume come 200 ml afterward. Including solutes come solvent can reason either expansion or convulsion (weird, right?) of the solution.

2. Molality (m)

Molality*, abbreviated by lower-case m, is the number of moles the solute divided by the variety of Kilograms that solvent. We say a solution is, because that example, a "3.0 molal solution."

This number ( → ) shows how to do an X-molal (X m) solution, wherein X is the number of moles the solute and n is the variety of Kilograms that solvent.

The volume that the solvent have the right to be used rather of the massive if its density is known. For example, in ~ 4˚C, 1 liter that H2O has a massive of 1 Kg. This measure up of concentration has actually a couple of advantages: (1) just a balance is required to prepare a systems of known concentration and also (2) variations of the thickness of the solvent v temperature (some can be significant) are irrelevant. Nevertheless, molality isn"t supplied that often.

*Note: In contemporary chemistry the term molality has fallen the end of use in donate of simply using the units: mol/Kg



Molality (m) is the number of moles that solute divided by the number of Kilograms of solvent. A 1 m solution includes 1 mole of solute for every 1 Kg the solvent. Usage of the unit mol/Kg is now preferred over molality.

Example 3

Calculate the molality the a equipment of MgCl2 the is developed by adding 998.2 g that water come 10.6 g of solid MgCl2.

Solution: Molality is mole of solute every Kg the solvent, therefore we"ll need those. An initial we"ll transform the fixed of solute come moles:

$$ \requirecancel \beginalign 10.6 \cancelg \, MgCl_2 &\left( \frac1 \, mol \, MgCl_295.2 \cancelg \, MgCl_2 \right) \\<5pt> &= 0.1113 \, mol \, MgCl_2 \endalign$$

Now convert the mass of the solvent come kilograms ("kilo" method 1000; there space 1000 grams in a kilogram)

$$ \requirecancel \beginalign 998.2 \cancelg \, H_2O &\left( \frac1 \, Kg1000 \cancelg \right) \\<5pt> &= 0.9982 \, Kg \, H_2O \endalign$$

Finally, the concentration is simply the proportion of those amounts:

$$\frac0.1113 \, mol \, MgCl_20.9982 \, Kg \, H_2O = 0.11 \, m \, MgCl_2$$

A note on molality

While molality can be quite helpful as a measurement of concentration, the isn"t too convenient for converting come molarity. The reason is that we typically don"t know exactly how much volume the solute is going to occupy in the solution. Part solutes even cause contraction the the solvent. For example, once 900 ml of distilled H2O is blended with 100 ml the ethanol (C2H5OH), the total volume the the resulting aqueous systems will be much less than 1 liter. The ethanol molecule are qualified of organizing H-bonded water molecule tightly roughly them, resulting in a smaller sized volume 보다 the combined volumes of the separate components.

3. Mole fraction ( χi )

The mole fraction is provided in part calculations because it is massless. The mole fraction of one ingredient of a solution is the variety of moles of that constituent separated by the total variety of moles that all contents of the solution. The amount of the mole fractions of all materials is one.

Mole fraction calculations occupational with any property the is proportional to the number of molecules (therefore moles) present, consisting of partial press for gases.

Mole fraction is an especially useful when we find out the fine details the the liquid state and also in statistical mechanics, the microscopic method to deriving thermodynamic characteristics of materials.

Mole fraction

The mole fraction ( χi ) of the ith ingredient of a mixture is the variety of moles of that component divided by the total number of moles of all components.

Example 4

Calculate the mole portion of both materials of a mixture the water (H2O) and ethanol (C2H5OH) the is 50% by massive in each component.

Solution: very first we require to uncover the number of moles of every component that the solution. To perform this you need the densities the water and ethanol. I looked them increase on Wikipedia, a pretty good source of chemistry properties.


Now it"s a an easy matter to uncover the 2 mole fractions. Mole fraction is usually offered the price χ, the Greek letter "chi."

XThe Greek alphabet

$$\chi_ethanol = \frac17.1317.13 + 55.55 = 0.236$$

$$\chi_H_2O = \frac55.5517.13 + 55.55 = 0.764$$

4. Components per ...

We commonly use devices like components per million (ppm) or components per trillion (ppt). The most typically used are

parts every million (ppm)

parts per billion (ppb)

parts per trillion (ppt)

These concentration dimensions are used commonly to describe low concentrations where low concentrations are significant, choose toxins. For example, the U.S. Food and Drug administration (USFDA) sets a limit on the allowable concentration that mercury (Hg) in food at 1 ppm because it is so toxic in very little amounts.

Example 5

Calculate the molarity of mercury (Hg) in ~ a concentration that 1 ppm in water.M

Wide equations, scroll L ↔ R

Solution: very first we ask "how plenty of moles the mercury is one atom the mercury?"

$$1 \cancelatom \, Hg \left( \frac1 \, mol \, Hg6.02 \times 10^23 \cancelatoms \right) = 1.66 \times 10^-24 \, moles$$

Now because that every million molecule of water, what is the volume of that water?

$$ \requirecancel 10^6 \cancelmolecules \, H_2O \left( \frac18 \cancelg \, H_2O6.02 \times 10^23 \cancelmolecules \right) \left( \frac1 \, together \, H_2O1000 \cancelg \, H_2O \right) = 3 \times 10^-20 \, L$$

The molarity is the variety of moles of Hg separated by the number of liters that solution. Notice that we"re in reality making one approximation here, namely that the enhancement of a little amount that mercury come water doesn"t significantly adjust its volume.

$$molarity = \frac1.66 \times 10^-24 \, mol3 \times 10^-20 \, L = 5.55 \times 10^-5 \, M$$

You deserve to see that 1 ppm is a very tiny concentration. Part substances space toxic at much smaller concentrations.

Practice problems


Calculate the molar concentration the a 415 ml solution containing 0.745 mole of HCl.


$$ \beginalign Molarity &= \frac\textmoles that solute\textliters of solution\\<5pt> &= \frac0.745 \, mol \, HCl0.415 \, L \\<5pt> &= 1.79 \, M \endalign$$


Calculate the molar concentration of one acetic mountain (CH3COOH) equipment containing 3.21 moles of HOAc in 4.50 liters. "HOAc" is a usual abbreviation for acetic acid.


$$ \beginalign Molarity &= \frac\textmoles of solute\textliters the solution\\<5pt> &= \frac3.21 \, mol \, HOAc4.50 \, L \\<5pt> &= 0.71 \, M \endalign$$


How plenty of moles that KI (potassium iodide) are existing in 125 ml the 0.5 M KI?


$$0.5 \, M \: \textmeans \: \frac0.5 \, mol \, KI1 \, L$$

Now set up a ratio to discover the number of moles that KI:

$$ \beginalign \frac0.5 \textmol KI1 \,L &= \fracx \textmol KI0.125 \, L \\<5pt> x(1) &= 0.5(0.125) \\<5pt> &= 0.062 \; \textmoles that KI \endalign$$


How countless liters the water are compelled to prepare a solution of 7.25 M MgCl2 native 4.89 moles of MgCl2 ?


$$7.25 \, M \: \textmeans \: \frac7.25 \, mol \, MgCl_21 \, L$$

Now set up a proportion to find the variety of liters that solution:

$$ \beginalign \frac7.25 \textmol MgCl_21 \,L &= \frac4.89 \textmol KIx \, L \\<5pt> 7.25 x &= 4.89(1) \\<5pt> &= 0.674 \; \textliters of solution \\<5pt> &= 674 \; \textmL \endalign$$


Calculate the molar concentration the a solution all set by including 34 g the NaCl (table salt) to 230 ml that H2O.


$$ \requirecancel \beginalign \textMolarity &= \frac\textmoles of solute\textliters of solution\\<5pt> &= \frac34 \cancelg \, NaCl \left( \frac1 \, mol \, NaCl54.45 \cancelg \, NaCl \right)0.230 \, L \\<5pt> &= 2.53 \, M \endalign$$

Be careful here. Molarity is mole of solute divided by liters the solution, no solvent. Below we"ve simply calculated an approximate molarity, however the volume impact of including a tiny amount of solute to water is typically small, for this reason this calculation most likely isn"t as well bad.


Calculate the concentration of a solution all set by dissolving 5.68 g the NaOH in sufficient water to make 400 ml of solution.


$$ \requirecancel \beginalign \textMolarity &= \frac\textmoles of solute\textliters of solution\\<5pt> &= \frac5.68 \cancelg \, NaOH \left( \frac1 \, mol \, NaOH40 \cancelg \, NaOH \right)0.400 \, L \\<5pt> &= 0.35 \, M \endalign$$

Be mindful here. Molarity is moles of solute split by liters the solution, not solvent. Below we"ve just calculated an approximate molarity, but the volume result of including a small amount of solute come water is typically small, therefore this calculation probably isn"t as well bad.


If a 2.34 g sample that dry ice (CO2) is dropped right into a sealed 500 ml bottle of orange KoolAid® and also the CO2 gas released dissolves fully in the drink, calculate the almost right molar concentration of CO2 in the newly-carbonated KoolAid®.


$$ \requirecancel \beginalign \textMolarity &= \frac\textmoles of solute\textliters of solution\\<5pt> &= \frac2.34 \cancelg \, CO_2 \left( \frac1 \, mol \, CO_244 \cancelg \, CO_2 \right)0.500 \, L \\<5pt> &= 0.11 \, M \endalign$$

This assumes that the CO2 doesn"t significantly readjust the volume that the solution. Gases can be solutes, too.


How many grams of beryllium chloride (BeCl2) are required to make 125 ml that a 0.050 M solution?


$$ \beginalign \text0.05 m means \: &\frac0.05 \, mol \, BeCl_21 \, together \, solvent \\<5pt> \frac0.05 \, mol \, BeCl_21 \, L &= \fracx \, mol \, BeCl_20.125 \, L \\<5pt> x &= 0.05(0.125) \\<5pt> &= 0.006 \, mol \, BeCl_2 \endalign$$

Then calculation the number of grams of BeCl2:

$$ \requirecancel 0.006 \cancelmol \, BeCl_2 \left( \frac80 \, g \, BeCl_21 \cancelmol \, BeCl_2 \right) = 0.05 \, g \, BeCl_2$$


How many grams that BeCl2 execute you need to add to 125 ml the water to make a 0.050 mol/Kg (molal) solution?


$$\textmolality = \frac\textmol solute\textKg solvent$$

The thickness of water is 1 g/mL or 1 Kg/L, so the mass of ours solvent is 0.125 Kg. Currently solve because that the variety of moles:

$$ \beginalign 0.05 \, m &= \fracx0.125 \\<5pt> x &= 0.05(0.125) \\<5pt> x &= 0.00625 \, mol \, BeCl_2 \endalign$$

Now calculate the mass of 0.00625 mole of BeCl2:

$$ \requirecancel \beginalign 0.00625 \cancelmol \, BeCl_2 &\left( \frac80 \, g \, BeCl_21 \cancelmol \, BeCl_2 \right) \\<5pt> &= 0.5 \, g \, BeCl_2 \endalign$$


The thickness of ethanol is 0.789 g/ml. How many grams the ethanol must be blended with 225 ml that water to make a 4.5% (volume/volume) mixture?


Remember that 4.5% is 4.5/100 or 0.045.

$$0.045 = \fracV_ethanol225 + V_ethanol$$

Now rearrange and solve for Vethanol:

$$ \beginalign 0.045 (225) + 0.045 \, V_ethanol &= V_ethanol \\<5pt> 0.955 \, V_ethanol &= 10.125 \\<5pt> V_ethanol &= 10.60 \, ml \endalign$$

Now use the density of ethanol to discover the mass:

$$ \requirecancel 10.60 \cancelml \left( \frac0.789 \, g1 \cancelml \right) = 8.36 \textg ethanol$$


Calculate the volume of a 0.50 M solution of calcium hydroxide (Ca(OH)2) if it has 25 g of Ca(OH)2.


Calculate the number of moles that Ca(OH)2:

$$ \requirecancel \beginalign 25 \cancelg \, Ca(OH)_2 &\left( \frac1 \, mol \, Ca(OH)_274.1 \cancelg \, Ca(OH)_2 \right) \\<5pt> &= 0.337 \, mol \, Ca(OH)_2 \endalign$$

Now set up a proportion to discover the volume in liters:

$$ \beginalign \frac0.5 \, mol \, Ca(OH)_21 \, L &= \frac0.337 \, mol \, Ca(OH)_2\bfx \, L \\<5pt> &= 0.674 \text together of solution \endalign$$


Calculate the mole fraction of sulfuric acid in a equipment made by adding 3.4 g of sulfuric mountain (H2SO4) come 3,500 ml of water.

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$$ \requirecancel \beginalign 3.4 \cancelg \, H_2SO_4 &\left( \frac1 \, mol \, H_2SO_498 \cancelg \, H_2SO_4 \right) \\<5pt> &= 0.0347 mol \, H_2SO_4 \endalign$$

$$3,500 \, ml \, H_2O \approx 3,500 \, g \, H_2O$$

$$ \requirecancel 3500 \cancelg \, H_2O \left( \frac1 \, mol \, H_2O18 \cancelg \, H_2O \right) = 194.4 \, mol \, H_2O$$

$$\chi_H_2SO_4 = \frac0.037194.4 + 0.037 = 1.9 \times 10^-4$$

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