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There's nothing more relaxing than a nice glass of lemonade on a hot summer day. But, did you know you are actually doing chemistry when you make it? The amount of lemonade powder you put into the glass, combined with the amount of water you put in to make the perfect concentration is molarity in action!

Definition of Molarity

Molarity is the concentration of solute dissolved in a solution expressed in units of moles per liter.

Molarity, or molar concentration, describes the concentration of an amount of a substance dissolved in a liquid. We call the substance we are dissolving a solute and the liquid is called a solvent. Specifically, molarity is defined by the number of moles per liter: mol/L. Solutes can consist of anything that dissolves into a liquid; they can be solids, other liquids, or even gases. You can learn more about them in our article on Solutions and Mixtures. If you know the amount of a solute in moles and the volume of solvent it is dissolved into, finding molarity is simple!

Molarity Equation

The standard molarity equation is thankfully very simple! It is :

The three variables are defined as:

  1. M is molar concentration expressed in mol/L
  2. n is the molar amount of the solute expressed in mol
  3. V is the volume of the solution expressed in L

How to find moles in molarity problems

Often, molarity problems won't just be as simple as dividing the moles of the solute by the liters of the solution. It is only one step in more complex problems. The beginning steps can involve many different things, but they will all lead to finally finding the amount of solute in moles and the volume in liters!

Instead of a problem just giving you moles, it may give you the number of total particles of the solute, the mass of solute used, or a reaction that creates the solute.

Let's take a look at a problem: it may seem complicated, but remember your end goal- you only need to find the total amount of moles of solute and the total volume of solution.

A student is preparing a nice bowl of soup, find the molarity of salt (NaCl) if this is the recipe:

1.5 liters of Water

60 grams of Salt

0.5 kg of Pasta

0.75 liters of Chicken Stock

200 grams of salted butter (3% salt by weight)

  1. Isolate the sources of solute aka. salt: 60g of Salt (100% Salt)200 grams of salted butter (3% salt)
  2. Find the molar mass of solute, which is salt in this example: Na (22.98 g/mol) + Cl (35.45 g/mol) = 58.44 g/mol
  3. Calculate moles of solute (salt) in pure salt: 60g/58.44g/mol = 1.027 mol
  4. find the weight of salt in butter:200g * 3% = 6g of NaCl
  5. Calculate moles of salt in butter:6g/58.44g/mol = 0.1027 mol
  6. add both sources of salt to find total moles:1.027+0.1027= 1.129 mol
  7. total all solvents used:1.5l+0.75l=2.25l of water
  8. divide moles of solute by liters of solvent:1.129 mol/2.25l= 0.501 M

Even though this problem was a lot of steps, as long as you keep your end goal in mind it is easy to work towards the solution! Always remember you need to find the total amount of solute and the total volume of solution.

If you run into any trouble following any of these steps, it might help to refresh your knowledge on moles and molar mass in general.

Uses of molarity

When reacting chemicals you almost always use solutions. In general, it is very hard to react two dry chemicals so one or both of your reactants must be in a solution. Just as it is with any chemical reaction moles are the key players, even if the reaction takes place in solution. So you probably need to calculate the mole ratios too. Luckily, these mole ratios don't even have to be calculated with moles, they can be directly calculated with molarity. Since molarity is always expressed with respect to a single liter, the mole ratio stays the same.

If you have the molarity of a solution and the volume of the solution it is very easy to calculate the moles in that solution. Just multiply both sides of the molarity equation by the volume to give you:

Let's use this equation in a simple precipitation reaction with two solutions

Pb(NO3)2(aq) + 2 KI(aq) → 2 KNO3(aq) + PbI2(s)

Using this reaction, find the volume of 1.2M KI(aq) solution required to create 1.5 moles of PbI2 if reacted with excess amounts of Pb(NO3)2(aq).

  1. Find the mole ratio of KI to PbI2:2 KI to make 1 PbI2
  2. Calculate the amount of KI needed:1.5mol of PbI2 multiplied by 2 equals 3mol of KI
  3. Calculate volume of solution needed:1.2M * 3 mol= 3.6L of KI(aq)

This problem is a simple example of how molarity is used in real chemical reactions. It is a critical component of almost every reaction

How to calculate dilutions using molarity

If you ever have to make a solution in the lab, or just want to pass your AP Chemistry exam, you are going to need to get used to molarities. One of the best uses of molarity is to calculate dilutions fast! In the lab, we usually only have a couple of solutions that are created at specific molarities. These solutions are called stock solutions.

A stock solution is a standardized solution of precisely known molar concentration that will be found in labs in large volumes

A stock solution of 2.0 M hydrochloric acid (HCl) is easy to produce and can be stored for a long time. Usually, however, you would need lower concentrations of HCl, think like 0.1 M or so, to do your reaction. In order to create this lower concentration solution, you must dilute the stock solution by adding more solvent. In some experiments such as titrations, low concentration acids and bases are more effective as they are easier to control. Thankfully there is an easy way to calculate the needed dilutions, just use this equation:

M1 & V1 refer to the volume and molarity of the stock solution, respectively. Usually, you will leave V1 as a variable as you are trying to find the volume of the solution you will need. V2 & M2 refer to the molarity and volume of the solution you are trying to make. Let's see an example to show how it would work in a lab:

When performing experiments, an independent variable will always have to change. Testing over a wide range of concentrations of a solution can show if the concentration has an impact on the dependent variable.

For an experiment, you want to test if the concentration of salt in water affects its ability to conduct electricity. To test this, you want to create solutions with molarities of 5M and 1M, each having 2L total. First, create a solution of 5M NaCl with solid salt, then create the 1M solution by diluting the 5M solution.

First, create the 5M solution,

Find the amount of salt in grams needed

Moles of salt will be 5M*2L=10mols required

Salt weighs 58.55g/mol, for 10mol multiply by 10, so 585.5g will be required

Add this amount of salt to 2L of water, resulting in the 5M solution.

Second, dilute the 5M solution to create 2L of 1M solution

Use the equation M1V1=M2V2


Solve for V1, (1M*2L)/5M=0.4L

Add 0.4L of the 5M to a beaker, then add enough water for the total volume to equal 2L. This means you will only have to add 1.6L of water. Remember, it is the total volume that needs to be 2L, not the amount of water you add.

So, to recap:

the first solution will need 585.5g of salt and 2L of water

the second solution will need 0.4L of the 5M solution and 1.6L of water

Molarity of Multiple Solutions Mixed

Sometimes you may end up having to find the concentration of two solutions after mixing them. It might seem complicated, but remember the steps to the original problem solving: 1st- find the total moles & 2nd- find the total volume!

Suppose you have multiple solutions with multiple volumes. You have to store this solution long term, but you only have one appropriate container for all of it. You decide to mix them all together but need to figure out the total volume and final molarity of it all.

Solution 1 is 3.0M and you have 0.5L of it.

Solution 2 is 1.5M and you have 0.75L of it

and Solution 3 is 0.75M and you have 1.0L of it

Find the final molarity after mixing all three solutions.

To start, you want to find the total moles present of solute that will be in the final mixture.

This is easily accomplished by adding up the moles of solute in each solution.

For Solution 1, this will be M1V1=n1


For Solution 2, this will be M2V2=n2


For Solution 3, this will be M3V3=n3


For the total it will be n1+n2+n3


Now, find the total volume which will be V1+V2+V3


Finally, like before, divide total moles by total volume


So from the example, it is easy to see what the equation should be when mixing any amount of solutions with the same solute. Divide the total moles by the total volume!

Total moles in the solution will be n1+n2+n3+..., but this will be M1V1+M2V2+M3V3+....

Total volume is simply V1+V2+V3+....

Dividing these leaves you with:

Molarity - Key takeaways

  • Remember the molarity equation: M=n/V
  • Understand how to manipulate the equation into the forms MV=n & V=n/M
  • Always identify the steps needed to find total moles of solute and total volume of solvent
  • Simplify hard problems into smaller steps to solve them

Frequently Asked Questions about Molarity

Molarity, or M,  is the concentration of solute dissolved in a solution expressed in units of moles per liter.

Molarity is the molar concentration of a solute. 

If there are 3 moles of salt, NaCl, dissolved in 1.5 liters of water, the molarity of salt is 2M (moles/liter).

To calculate molarity, divide the total amount of solute in moles by the total amount of solution in liters. M=n/V

The molarity equation for a mixture of solutions with the same solute is Msolution=(M1V1+M2V2+...)/(V1+V2+...).

The equation for finding molarity is to divide the total amount of solute in moles by the total amount of solution in liters. M=n/V

Final Molarity Quiz


What is Molarity?

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Molarity is the molar concentration of a solute in a solution expressed in units of mol/L.

Show question


What is dilution?

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Dilution is reducing the concentration of a solution by adding more of the same solvent.

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What is a mole?

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A mole is a measure of the amount of units of a substance. One mole is 6.022x1023 units of a substance.

Show question


What is the molarity equation?

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Show question


What is the dilution equation?

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Show question


What is a stock solution

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A stock solution is a concentrated, stable solution with a very well known & precise concentration that is often diluted. 

Show question


What is your goal when solving complex molarity problems?

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Always try to find the total sum of moles of solute from all sources and the total volume of solution from all sources.

Show question


What is the molarity of a solution with 0.75 moles of solute and 12 liters of solution?

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Show question


What is the molarity if 55g of NaCl (MW=58.44g/mol) and 3.011x1023 molecules of NaCl are added to 2 liters of water?

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0.72 M NaCl

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What is the final molarity if you mix 2L of a 5M solution with 4L of a 6M solution?

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How much salt (NaCl, MW=58.44g/mol) will you have to add to water to make 0.75L of 2M solution?

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1.5mol or 87.66g

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How many moles of solute are there in 0.7L of a 1.3M solution?

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0.91 mol

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What is the final molarity if you add 1 mol of solute to 500ml of a 2M solution? (Hint: Increasing concentration)

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Total moles: 1 mol + (0.5L*2M)= 2 mol

Total volume: 0.5L

Show question


How much solvent do I have to add to 5L of 5M HCl to dilute it to 3M?

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Show question


What is the molarity of a solution with density 1kg per liter if it is labelled as 10% NaCl by weight?

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Assume 1L of Solution

1kg of solution times 10% is 100g of NaCl

100g/58.44(g/mol)=1.711 mol of NaCl in 1 liter

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