In this section, we"re largely working via aqueous (water-as-solvent) options, yet the very same procedures of concentration can apply to any kind of other solvent. When the solven is not offered, or if we recognize the solutes as water-soluble (solutes many typically dissoved in water), we assume that the solution is aqueous.

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Strength of a solution

We need a method to represent the relative amount of solute liquified in a solvent.

It provides feeling that some options are more powerful than others: including a tiny little bit of acid to a gallon of water, for example, could not even be detectable. But include a lot of acid and also you have a dangerous solution that requirements to be tackled closely.

We need a way to represent the loved one toughness of a solution. We"ll call it concentration.

There are numerous approaches of concentration measurement, each offered in different kinds of cases.


If we liquified simply a couple of crystals of table salt (NaCl) in water, we might not even have the ability to taste it. But if we liquified a bunch of NaCl in the same amount of water (and also you might be surprised how a lot will certainly dissolve), it would certainly taste exceptionally salty.

On a relative range, one solution is dilute and the other is concentrated. We"d favor to be able to put this on some sort of numerical scale so that we can say just precisely just how dilute or focused a solution is.

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

All of the concentration measurement approaches extended listed below consist of some measure of the amount of solute (in grams, moles or atoms/molecules) divided by the amount of solvent (in devices of mass, volume, moles or variety of atoms/molecules).

1. Molarity (M)

Molarity, abbreviated (M) is most likely the a lot of commonly supplied meacertain of solution concentration. Molarity is the variety of moles of solvent split by the variety of liters of solution. We describe a solution, for instance, as a "1.5 molar solution" or "1.5 M".

Note that we divide by the total variety of liters of solution, consisting of the solute, not the variety of liters of solvent to which the solute was added — it"s a vital (if frequently small) difference.

Practical tip:

This figure ( → ) mirrors exactly how to make an X-molar (X M) solution, wbelow X is the wanted molar concentration. We have a targain volume (1 liter in this instance, yet it could be anything). The solute is dissolved in a smaller sized volume of solvent, then the complete volume is adjusted to the final desired amount.



Molarity (M) is the number of moles of solute divided by the total volume of the solution in liters. A 1 M solution has 1 mole of solute for eexceptionally 1 L of solution.

Example 1

Calculate the molarity of a NaCl solution developed by disfixing 62 g of NaCl in water and adjusting the total volume to 0.50 liters.

Solution: Begin by calculating the variety of moles of solute


Then divide it by the full number of liters of solution, 0.5 L in this case:


Example 2

How many kind of grams of (NH4)2SO4 (ammonium sulfate) should be included to water to make 200 ml of a 1.6 M solution?

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


Then convert it to grams.


Note that in making this solution, we"d desire to dissolve the solute in less than 200 ml of water, then lug the full volume to 200 ml afterward. Adding solutes to solvent have the right to reason either expansion or contractivity (weird, right?) of the solution.

2. Molality (m)

Molality*, abbreviated by lower-situation m, is the number of moles of solute divided by the variety of Kilograms of solvent. We say a solution is, for instance, a "3.0 molal solution."

This number ( → ) mirrors just how to make an X-molal (X m) solution, wbelow X is the variety of moles of solute and also n is the variety of Kilograms of solvent.

The volume of the solvent have the right to be supplied rather of the mass if its thickness is recognized. For instance, at 4˚C, 1 liter of H2O has actually a mass of 1 Kg. This meacertain of concentration has a pair of advantages: (1) just a balance is compelled to prepare a solution of well-known concentration and (2) variations of the thickness of the solvent with temperature (some can be significant) are irrelevant. Nevertheless, molality isn"t provided that often.

*Note: In modern chemistry the term molality has fallen out of usage in favor of simply utilizing the units: mol/Kg



Molality (m) is the number of moles of solute separated by the number of Kilograms of solvent. A 1 m solution includes 1 mole of solute for eincredibly 1 Kg of solvent. Use of the unit mol/Kg is now wanted over molality.

Example 3

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

Solution: Molality is moles of solute per Kg of solvent, so we"ll require those. First we"ll convert the mass of solute to moles:

$$ equirecancel eginalign 10.6 cancelg , MgCl_2 &left( frac1 , mol , MgCl_295.2 cancelg , MgCl_2 ight) \<5pt> &= 0.1113 , mol , MgCl_2 endalign$$

Now transform the mass of the solvent to kilograms ("kilo" implies 1000; there are 1000 grams in a kilogram)

$$ equirecancel eginalign 998.2 cancelg , H_2O &left( frac1 , Kg1000 cancelg ight) \<5pt> &= 0.9982 , Kg , H_2O endalign$$

Finally, the concentration is just the ratio of those amounts:

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

A note on molality

While molality have the right to be rather beneficial as a measurement of concentration, it isn"t also convenient for converting to molarity. The factor is that we generally don"t recognize how much volume the solute is going to occupy in the solution. Some solutes also cause contractivity of the solvent. For instance, as soon as 900 ml of distilled H2O is mixed with 100 ml of ethanol (C2H5OH), the total volume of the resulting aqueous solution will be much less than 1 liter. The ethanol molecules are qualified of organizing H-bonded water molecules tightly approximately them, resulting in a smaller volume than the combined quantities of the separate components.

3. Mole fraction ( χi )

The mole fraction is offered in some calculations bereason it is massmuch less. The mole fractivity of one constituent of a solution is the variety of moles of that constituent split by the full variety of moles of all components of the solution. The sum of the mole fractions of all components is one.

Mole fraction calculations job-related with any kind of home that is proportional to the variety of molecules (therefore moles) existing, consisting of partial press for gases.

Mole fraction is especially advantageous when we learn the fine details of the liquid state and also in statistical mechanics, the microscopic strategy to deriving thermodynamic characteristics of materials.

Mole fraction

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

Example 4

Calculate the mole fraction of both components of a mixture of water (H2O) and ethanol (C2H5OH) that is 50% by mass in each component.

Solution: First we should find the variety of moles of each component of the solution. To do this you need the densities of water and also ethanol. I looked them up on Wikipedia, a pretty good source of chemical properties.


Now it"s a simple issue to discover the two mole fractions. Mole fraction is typically given the symbol 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. Parts per ...

We commonly usage devices favor components per million (ppm) or parts per trillion (ppt). The a lot of commonly provided are

parts per million (ppm)

parts per billion (ppb)

parts per trillion (ppt)

These concentration dimensions are supplied typically to define low concentrations where low concentrations are considerable, choose toxins. For instance, the UNITED STATE Food And Drug Administration (USFDA) sets a limit on the allowable concentration of mercury (Hg) in food at 1 ppm bereason it is so toxic in incredibly small quantities.

Example 5

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

Wide equations, scroll L ↔ R

Solution: First we ask "exactly how many moles of mercury is one atom of mercury?"

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

Now for every million molecules of water, what is the volume of that water?

$$ equirecancel 10^6 cancelmolecules , H_2O left( frac18 cancelg , H_2O6.02 imes 10^23 cancelmolecules ight) left( frac1 , L , H_2O1000 cancelg , H_2O ight) = 3 imes 10^-20 , L$$

The molarity is the number of moles of Hg divided by the number of liters of solution. Notice that we"re actually making an approximation right here, namely that the enhancement of a little amount of mercury to water does not considerably readjust its volume.

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

You deserve to watch that 1 ppm is a really small concentration. Some substances are toxic at much smaller concentrations.

Practice problems


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


$$ eginalign Molarity &= frac extmoles of solute extliters of solution\<5pt> &= frac0.745 , mol , HCl0.415 , L \<5pt> &= 1.79 , M endalign$$


Calculate the molar concentration of an acetic acid (CH3COOH) solution containing 3.21 moles of HOAc in 4.50 liters. "HOAc" is a prevalent abbreviation for acetic acid.


$$ eginalign Molarity &= frac extmoles of solute extliters of solution\<5pt> &= frac3.21 , mol , HOAc4.50 , L \<5pt> &= 0.71 , M endalign$$


How many kind of moles of KI (potassium iodide) are current in 125 ml of 0.5 M KI?


$$0.5 , M : extmeans : frac0.5 , mol , KI1 , L$$

Now erected a propercent to find the number of moles of KI:

$$ eginalign frac0.5 extmol KI1 ,L &= fracx extmol KI0.125 , L \<5pt> x(1) &= 0.5(0.125) \<5pt> &= 0.062 ; extmoles of KI endalign$$


How many liters of water are forced to prepare a solution of 7.25 M MgCl2 from 4.89 moles of MgCl2 ?


$$7.25 , M : extmeans : frac7.25 , mol , MgCl_21 , L$$

Now erected a propercentage to discover the number of liters of solution:

$$ eginalign frac7.25 extmol MgCl_21 ,L &= frac4.89 extmol KIx , L \<5pt> 7.25 x &= 4.89(1) \<5pt> &= 0.674 ; extliters of solution \<5pt> &= 674 ; extmL endalign$$


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


$$ equirecancel eginalign extMolarity &= frac extmoles of solute extliters of solution\<5pt> &= frac34 cancelg , NaCl left( frac1 , mol , NaCl54.45 cancelg , NaCl ight)0.230 , L \<5pt> &= 2.53 , M endalign$$

Be careful below. Molarity is moles of solute separated by liters of solution, not solvent. Here we"ve simply calculated an approximate molarity, however the volume effect of including a small amount of solute to water is typically tiny, so this calculation most likely isn"t also bad.


Calculate the concentration of a solution ready by disresolving 5.68 g of NaOH in enough water to make 400 ml of solution.


$$ equirecancel eginalign extMolarity &= frac extmoles of solute extliters of solution\<5pt> &= frac5.68 cancelg , NaOH left( frac1 , mol , NaOH40 cancelg , NaOH ight)0.400 , L \<5pt> &= 0.35 , M endalign$$

Be mindful below. Molarity is moles of solute split by liters of solution, not solvent. Here we"ve simply calculated an approximate molarity, however the volume impact of including a little amount of solute to water is generally small, so this calculation more than likely isn"t also bad.


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


$$ equirecancel eginalign extMolarity &= frac extmoles of solute extliters of solution\<5pt> &= frac2.34 cancelg , CO_2 left( frac1 , mol , CO_244 cancelg , CO_2 ight)0.500 , L \<5pt> &= 0.11 , M endalign$$

This assumes that the CO2 doesn"t substantially readjust the volume of the solution. Gases deserve to be solutes, also.


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


$$ eginalign ext0.05 m means : &frac0.05 , mol , BeCl_21 , L , solvent \<5pt> frac0.05 , mol , BeCl_21 , L &= fracx , mol , BeCl_20.125 , L \<5pt> x &= 0.05(0.125) \<5pt> &= 0.006 , mol , BeCl_2 endalign$$

Then calculate the variety of grams of BeCl2:

$$ equirecancel 0.006 cancelmol , BeCl_2 left( frac80 , g , BeCl_21 cancelmol , BeCl_2 ight) = 0.05 , g , BeCl_2$$


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


$$ extmolality = frac extmol solute extKg solvent$$

The thickness of water is 1 g/mL or 1 Kg/L, so the mass of our solvent is 0.125 Kg. Now fix for the number of moles:

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

Now calculate the mass of 0.00625 moles of BeCl2:

$$ equirecancel eginalign 0.00625 cancelmol , BeCl_2 &left( frac80 , g , BeCl_21 cancelmol , BeCl_2 ight) \<5pt> &= 0.5 , g , BeCl_2 endalign$$


The density of ethanol is 0.789 g/ml. How many kind of grams of ethanol must be mixed through 225 ml of 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 rearselection and also fix for Vethanol:

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

Now usage the density of ethanol to discover the mass:

$$ equirecancel 10.60 cancelml left( frac0.789 , g1 cancelml ight) = 8.36 extg ethanol$$


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


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

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

Now put up a proportion to uncover the volume in liters:

$$ eginalign frac0.5 , mol , Ca(OH)_21 , L &= frac0.337 , mol , Ca(OH)_2fx , L \<5pt> &= 0.674 ext L of solution endalign$$


Calculate the mole fraction of sulfuric acid in a solution made by including 3.4 g of sulfuric acid (H2SO4) to 3,500 ml of water.

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$$ equirecancel eginalign 3.4 cancelg , H_2SO_4 &left( frac1 , mol , H_2SO_498 cancelg , H_2SO_4 ight) \<5pt> &= 0.0347 mol , H_2SO_4 endalign$$

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

$$ equirecancel 3500 cancelg , H_2O left( frac1 , mol , H_2O18 cancelg , H_2O ight) = 194.4 , mol , H_2O$$

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

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