In this section, we"re largely functioning through aqueous (water-as-solvent) remedies, but the very same procedures of concentration have the right to apply to any kind of various other solvent. When the solven is not provided, or if we identify 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 way to reexisting the family member amount of solute liquified in a solvent.

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

We require a method to reexisting the loved one strength of a solution. We"ll contact it concentration.

There are numerous techniques of concentration measurement, each provided in various kinds of cases.


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

On a family member range, one solution is dilute and also the other is concentrated. We"d choose to be able to put this on some sort of numerical scale so that we could say just exactly just how dilute or concentrated a solution is.

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

All of the concentration measurement approaches extended listed below consist of some meacertain of the amount of solute (in grams, moles or atoms/molecules) split by the amount of solvent (in units of mass, volume, moles or number of atoms/molecules).

1. Molarity (M)

Molarity, abbreviated (M) is probably the the majority of typically used meacertain of solution concentration. Molarity is the variety of moles of solvent separated by the number of liters of solution. We refer to a solution, for example, as a "1.5 molar solution" or "1.5 M".

Keep in mind that we divide by the complete number of liters of solution, including the solute, not the number of liters of solvent to which the solute was added — it"s an essential (if regularly small) difference.

Practical tip:

This figure ( → ) mirrors exactly how to make an X-molar (X M) solution, where X is the wanted molar concentration. We have a tarobtain volume (1 liter in this situation, but it might be anything). The solute is dissolved in a smaller sized volume of solvent, then the full volume is adjusted to the last preferred amount.



Molarity (M) is the number of moles of solute separated by the complete volume of the solution in liters. A 1 M solution has 1 mole of solute for every 1 L of solution.

Example 1

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

Solution: Begin by calculating the variety of moles of solute


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


Example 2

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

In this problem we know the concentration, we simply have to understand the amount of solute essential to attain it. Fist uncover that number of moles:


Then convert it to grams.


Note that in making this solution, we"d want to disresolve the solute in much less than 200 ml of water, then carry the complete volume to 200 ml afterward. Adding solutes to solvent have the right to cause either development or contraction (weird, right?) of the solution.

2. Molality (m)

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

This figure ( → ) mirrors how to make an X-molal (X m) solution, wright here X is the number of moles of solute and n is the number of Kilograms of solvent.

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

*Note: In modern-day chemistry the term molality has fallen out of use in favor of just using 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 consists of 1 mole of solute for every 1 Kg of solvent. Use of the unit mol/Kg is currently desired over molality.

Example 3

Calculate the molality of a solution of MgCl2 that is developed by including 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 convert the mass of the solvent to kilograms ("kilo" means 1000; tright here 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 deserve to be fairly helpful as a measurement of concentration, it isn"t too convenient for converting to molarity. The reason is that we commonly do not recognize just how much volume the solute is going to occupy in the solution. Some solutes also cause contractivity of the solvent. For instance, once 900 ml of distilled H2O is blended through 100 ml of ethanol (C2H5OH), the full volume of the resulting aqueous solution will certainly be much less than 1 liter. The ethanol molecules are qualified of organizing H-bonded water molecules tightly approximately them, leading to a smaller volume than the merged quantities of the sepaprice components.

3. Mole fraction ( χi )

The mole fraction is supplied in some calculations bereason it is massless. The mole fraction of one constituent of a solution is the number of moles of that constituent split by the complete number of moles of all components of the solution. The amount of the mole fractions of all components is one.

Mole fractivity calculations work with any type of home that is proportional to the number of molecules (therefore moles) present, including partial pressure for gases.

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

Mole fraction

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

Example 4

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

Solution: First we should uncover the number of moles of each component of the solution. To execute this you require the densities of water and also ethanol. I looked them up on Wikipedia, a pretty great source of chemical properties.


Now it"s a simple issue to uncover the two mole fractions. Mole fraction is normally 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 generally use devices choose components per million (ppm) or parts per trillion (ppt). The many generally used are

parts per million (ppm)

parts per billion (ppb)

components per trillion (ppt)

These concentration dimensions are offered generally to describe low concentrations where low concentrations are considerable, favor toxins. For example, the UNITED STATE Food And Drug Administration (USFDA) sets a limit on the allowable concentration of mercury (Hg) in food at 1 ppm because it is so toxic in exceptionally tiny 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 "just how many type of 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 eincredibly 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 variety of moles of Hg separated by the variety of liters of solution. Notice that we"re actually making an approximation below, namely that the addition of a little amount of mercury to water doesn"t significantly readjust its volume.

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

You deserve to check out that 1 ppm is an extremely little 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 widespread 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 type of moles of KI (potassium iodide) are existing in 125 ml of 0.5 M KI?


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

Now erected a propercentage 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 required 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 put up a propercentage to discover the variety 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 ready 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 mindful below. Molarity is moles of solute separated by liters of solution, not solvent. Here we"ve just calculated an approximate molarity, yet the volume impact of including a tiny amount of solute to water is commonly little, so this calculation most likely isn"t also bad.


Calculate the concentration of a solution ready by disfixing 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 careful below. Molarity is moles of solute separated by liters of solution, not solvent. Here we"ve simply calculated an approximate molarity, yet the volume impact of adding a small amount of solute to water is normally tiny, so this calculation probably isn"t as well bad.


If a 2.34 g sample of dry ice (CO2) is dropped right into a sealed 500 ml bottle of ovariety KoolAid® and the CO2 gas released dissolves entirely 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 does not substantially change the volume of the solution. Gases have the right to be solutes, as well.


How many type of grams of beryllium chloride (BeCl2) are needed 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 number 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 kind of grams of BeCl2 execute you must include to 125 ml of water to make a 0.050 mol/Kg (molal) solution?


$$ extmolality = frac extmol solute extKg solvent$$

The density 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 thickness of ethanol is 0.789 g/ml. How many grams of ethanol need to be combined with 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 rearrange 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 thickness of ethanol to uncover 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 variety 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 set up a propercentage to find 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 fractivity of sulfuric acid in a solution made by including 3.4 g of sulfuric acid (H2SO4) to 3,500 ml of water.

See more: 5. What Is The Percent Error For Your Experimental Determination Of Avogadro’S Number?


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

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

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

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

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