If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Molarity

AP.Chem:
SPQ‑3 (EU)
,
SPQ‑3.A (LO)
,
SPQ‑3.A.2 (EK)
The most common way to express solution concentration is molarity (M), which is defined as the amount of solute in moles divided by the volume of solution in liters: M = moles of solute/liters of solution. A solution that is 1.00 molar (written 1.00 M) contains 1.00 mole of solute for every liter of solution. Created by Sal Khan.

Want to join the conversation?

  • piceratops tree style avatar for user ŇØŦ€ŞŁΔ¥€Ř
    Is molarity the same as concentration?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      Concentration in general is a measure of how abundant a component is in a mixture. And molarity is one way to measure concentration and is the most commonly used in chemistry. But there are other ways to measure it too like molality, weight %, ppm (parts per million), etc. So for most chemistry purposes, molarity is the same as concentration since it's the primary unit we use.

      Hope that helps.
      (9 votes)
  • blobby green style avatar for user grantbailey
    Is there such thing as a maximum molarity a solution could attain? And why is this?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      Yes, a solution can become saturated. When you dissolve a solute into a solvent you create a solution where the solute has a concentration measured in molarity. If you can no longer dissolve any more solute into the solvent, your solution has become saturated therefore essentially capping the molarity of the solute. When you dissolve something like salt, sodium chloride or NaCl, into water, the water molecules surround each individual sodium and chloride ion thereby turning them into aqueous ions and increasing the molarity. If you keep adding more salt you need more and more water molecules to surround those ions to actually make them dissolve. Eventually you run out of free water molecules to surround the ions and the salt can no longer dissolve so it stays in its unaltered solid. The solid form does not add to the molarity since molarity only measures aqueous species in solution. You would observe solid salt particles just falling to the bottom of the water in a saturated solution. Hope that helps.
      (4 votes)
  • primosaur seedling style avatar for user Agnes Gwanbobmuga
    At , how do I figure out the molar mass
    (2 votes)
    Default Khan Academy avatar avatar for user
  • duskpin seedling style avatar for user Kat
    How do you find the mass of a solute?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      I'm going to assume that you mean 'how do you find the mass of the solute if you're only given molarity' since that's the title of the video.

      So molarity is defined as moles of solute in moles divided by volume of solution in liters. Or as an equation M = n/V, where M is molarity, n is moles of solute, and V is volume. If you multiply the equation by volume you have a way to solve for number of moles, or MV = n. So basically you can multiply the molarity by volume to find the number of moles of solute. Then you can use the molar mass of the solute to convert moles of solute into grams of solute, or the mass.

      Hope that helps.
      (2 votes)
  • duskpin ultimate style avatar for user Shriya
    So, here is the thing: I'm an 8th grader placed in an advanced class and we learnt this the past Friday. I get how to find the molarity, but not "How many grams of Beryllium Chloride are needed to make 125 mL of a 0.050 M solution?" My teacher did go over it, but I'm still stuck. I also don't get the question "What will the volume of a 0.050 M Solution be if it contains 25 grams of calcium Hydroxide?" It would be really appreciated if someone could help me:)
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      So this is a stoichiometry problem where we have figure out amounts of chemicals. For these problems it is invaluable to be well acquainted with something called dimensional analysis. Dimensional analysis is a way of converting quantities from one unit to another and takes all the guess work out of it. Essentially it amounts to performing multiple multiplications/divisions where every step you are cross canceling units until you arrive at the desired unit.

      For the first question involving beryllium chloride we can use dimensional analysis to quickly solve the problem. We begin with volume of a solution in milliliters and its concentration in molarity, and we ultimately want the mass in grams. So we're going to either multiply or divide the volume and the molarity to get grams. If we look at just the units we know which to do. Molarity is a unit of concentration defined as moles/liter. In order for the units to cancel we need the volume to be in liters instead of milliliters. So we need to convert 125 mL to liters. We know there are 1000 mL in 1 liter so the conversion ratio is 1: 1000. So for the mL units to cancel and give us liters as the unit we need to do: 125 mL x (1 L/1000 mL) = 0.125 L. So I divided by 1000 essentially which is the same as multiplying by its reciprocal 1/1000. Now this volume can be multiplied by the molarity to yield us moles of beryllium chloride: 0.125 L x (0.050 mol/ 1 L) = 0.00625 mol. Again see how the units cross cancel to give us a new unit. To convert moles of a chemical to grams we need a conversion ratio which uses moles and grams: the molar mass of beryllium chloride. We can get this number by adding the relative atomic masses of both elements on the periodic table: 9.0122 (Be) + 35.45 (Cl) = 45.4622 g/mol. Again the molar mass of a chemical says how many grams are contained within 1 mole of that chemical. So we can take our mole number from before and multiply it by the molar mass: 0.00625 mol x (45.4622 g/1 mol) = 0.27788875 g. Accounting for sig figs the answer is properly reported as 0.28 grams of beryllium chloride.

      The second question involving calcium hydroxide is also solved easily using dimensional analysis. We have the mass of calcium hydroxide in grams and the concentration of the solution in molarity, which has units of mol/L again, and we want the volume of the solution. So we need to multiply the mass by the molarity to get the volume, but the units have to agree first so we need to convert the grams into moles. Again we'll use the molar mass, but of calcium hydroxide. This one is a little different and we need to know the chemical formula for calcium hydroxide, which is Ca(OH)2. The formula tells us there is one calcium atom, two hydrogen atoms, and two oxygen atoms. So we need their relative atomic masses from the periodic table which gives us: 40.078 (Ca) + (2 x 1.008) (H) + (2 x 15.999) (O) = 74.092 g/mol. So now multiplying the mass in grams by the molar mass of calcium hydroxide we get the moles: 25 g x (1 mol/ 74.092 g) = 0.33741834476 mol of calcium hydroxide. Now multiplying this mole number by the molarity yields the volume of the solution in liters: 0.33741834476 mol x (1 L/0.050 mol) = 6.74836689521 L. Accounting for the sig figs the answer is properly reported as 6.7 L.

      Hope that helps.
      (2 votes)
  • blobby green style avatar for user nicey429
    how do you calculate mole
    (1 vote)
    Default Khan Academy avatar avatar for user
  • winston default style avatar for user PokemonsterBlue2
    What do we do if the grams of the solution is not given?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Robot
    What is the mass percent of potassium chloride (KCl) in a solution with a concentration of 2.24 M potassium chloride if its density is 1.131 g/mL?
    (0 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user algifarihaikal123
    i still dont get a clear idea of sig fig. can anybody explain it please...
    (0 votes)
    Default Khan Academy avatar avatar for user
    • female robot ada style avatar for user SarahLDeMonia
      Hi there! Significant figures can be a bit tricky but they become much easier with examples. For example, the numbers 0.00451, 2.30, and 4,500.
      Let's look at 0.00451. Because of the leading zeroes rule, you would start counting significant digits starting at the '4'. Thus, this number has 3 significant digits. Once again, this is because leading zeroes are not significant.
      As for 2.30, you notice there is a zero but it is after the decimal place. Because it was deliberately included by the researcher it is significant. Thus, we would say that 2.30 has 3 significant digits.
      As for the last example, 4,500, you notice there are zeroes that follow the five but because there is no decimal place we consider these non-significant. A neat trick is to see how the number looks in scientific notation. 4,500 can be written as 4.5 x 10^3. In other words, the zeroes are not needed to convey the significance of the number. If I had 4,500.0, however, the researcher deliberately was trying to say that the tenths place was significant and so they left the decimal. In 4,500.0, then, there would be 5 significant digits.
      I found a response by Chris Martin on Quora which does a better job of explaining this concept:
      "Because they are assumed to be rounded zeroes instead of measured or calculated ones. The existence of a decimal point indicates that the zeroes are part of the actual measured or calculated value."
      I hope this helps you! Let me know if you have any more questions!
      (1 vote)

Video transcript

- [Instructor] In this video, we're gonna talk about one of the most common ways to measure solute concentration in a solution. And that is molarity. And molarity is defined as the number of moles of solute, the thing that we are dissolving in a solvent, that divided by the liters of solution. So let's just do an example and see if we can calculate the molarity of a solution. So let's say that I have, this is a container here. And I'm going to dissolve some sodium sulfate in water. So sodium sulfate is the solute and water is the solvent. Together, they give us this solution. And let's say we have a total, the total volume of solution is 250 milliliters of solution. And that solution is made up of just to give ourselves a bit of a refresher. We have the solvent, which is H2O, it is water in this situation. And you might say, do we have 250 milliliters of water? And the answer would be not quite because the 250 milliliters that's the volume of the water plus the sodium sulfate. And so we're gonna have some sodium sulfate in here. And let's say we know that we have 35.5 grams of sodium sulfate. That is the formula for sodium sulfate. And so given this information, how do we figure out molarity? Well, the first thing you might say is okay, I know the number of grams of sodium sulfate. I need to figure out the number of moles. And to figure out the number of moles, you'd have to look at the molar mass. You could figure that out from a periodic table of elements. But just to speed us along, I will help you out a little bit here. The molar mass sodium sulfate, molar mass is 142.04 grams per mole. So given everything I've now told you, see if you can pause this video and figure out the molarity of this solution. What's the molarity of the sodium sulfate in this solution? All right, now let's work through this together. So first, we wanna figure out the number of moles of solute. So we can start with the mass of solute that we have right over here. So we have 35.5 grams of sodium sulfate. And now if we wanna figure out the number of moles, see I'm gonna multiply this times something that would cancel out the grams. And so I don't want grams per mole. I want moles per gram. And so I could write this. I could multiply this times for every one mole of sodium sulfate, NA2SO4, we have 142.04. I'll write this way, grams of sodium sulfate. And you can see very clearly that that will cancel with that and we're left with moles of sodium sulfate. So we'll get a calculator out in a second and just take 35.5 and divide that by 142.04. And then to figure out molarity, we wanna divide by the liters of solution. So up here we have a calculation for number of moles and then the liters have solution, 250 milliliters is the same thing as 0.250 liters of our solution. And now we can just use our calculator to figure out what this is. 35.5 divided by 142.04 equals that, and then we divide that by 0.25. I could just throw a zero in there. And then that gets us that right over there. And then we can think about how many significant figures we have. So we have three over here. We have one, two, three, four, five over here. We have three over here. So, I would say that we have three significant figures. And so we would round this right most or this nine right over here. Well, if we round that up, we get 1.00, if we were to go to three significant figures. So that gets us 1.00. And then you might say what is the units here? And what people will normally say is this is 1.00 molar. When you see this capital M right over here, that is the unit for more molarity but they're really talking about the number of moles of solute per liter of solution. And we are done.