Molarity Made Easy: How to Calculate Molarity and Make Solutions

By Stevie Adams / in , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , /

Welcome to Molarity Made Easy,
brought to you by Ketzbook In this video,
we are going to learn how to do molarity calculations and how to make a solution of a certain molarity. But first, what is molarity? As the name suggests,
molarity has something to do with moles, and we remember that the mole is a scientist’s
way of counting the number of particles there are Molarity is a concentration measurement that
uses moles. It measures how concentrated a solution is
based on the number of solute particles dissolved. The formula of molarity is M equals n over
V, where M is molarity,
n is the number of moles of solute, and V is the volume of the solution in liters. Let’s go ahead and try a sample problem. What is the molarity of a 125 mL solution
containing .05 moles of hydrochloric acid. Before we start plugging numbers in our calculator,
it’s good to identify what we are solving for and the other variables in the question. The question starts out, “what is the molarity,”
which means we are solving for molarity. 125 mL is the volume, and .05 is the number
of moles. Next, write down the appropriate equation,
which is M equals n over V. Because M is what we are solving for,
we do not need to rearrange the equation. Substitute .05 moles in for n, and divide
that by the volume. You may be tempted to write 125 mL in the
denominator, but that would be incorrect because V should
be the volume in liters, not mL. In order to convert 125 mL to liters, you can
multiply it by a conversion factor fraction. Because one liter equals 1000 mL,
we put 1000 mL on the bottom and one liter on the top. The volume works out to be 0.125 L.
However, because we are dealing with metric prefixes,
we could have done the conversion simply by moving the decimal point according to the
prefix. For milli, just move the decimal point three
spots to the left. Back to the original problem,
V equals 0.125 L, so put that in the denominator. Do the division, and the answer works out
to be 0.40 moles per liter. In general, the units of molarity are moles
per liter. Let’s try another problem where we are solving
for something different. How many moles of sodium hydroxide are in
38 mL of 0.5 mol/L sodium hydroxide? The problem is asking us to calculate moles,
and it tells us that the volume is 38 mL. However, for molarity we will need to convert
that to liters, which we can do by simply moving the decimal
point three places to the left. 0.50 mol/L is the molarity. Next, write down the appropriate equation. M equals n over V. Because we are solving for n,
we multiply both sides of the equation by V.
On the right side, V over V cancels out, leaving n all by itself. We can rewrite the equation as n equals M
times V. Plug in the values for M and V remembering
that V has to be in liters. 0.5 mol/L times 0.038 liters equals 0.019. As for the units, liters on the top and bottom
cancel, and the answer has the units of moles. Let’s try another one. How much 2.5 mol/L sulfuric acid should you
use if you need 0.12 moles of sulfuric acid? This problem simply asks us “how much sulfuric
acid,” which is a little vague,
but think about what you would do in a lab. How would you measure out a solution of sulfuric
acid? By its volume, of course,
so the problem is asking us to solve for volume. 2.5 mol/L is the molarity, and 0.12 moles
is n. Next, write down the appropriate equation. We use n equals M times V because it is closer
to the form of the equation we need. It order to get V all by itself,
divide both sides of the equation by M. On the right side, M over M cancels out,
leaving V all by itself. We can rewrite the equation as V equals n
over M. Finally, plug in the values for n and M.
0.12 moles divided by 2.5 mol/L equals 0.048. As for the units, moles on the top and bottom
cancel, leaving us with liters as the units. Because the number is less than one,
we convert it to mL by moving the decimal point three places to the right,
giving us the final answer of 48 mL. Now for our final problem,
let’s talk about how you would actually make a solution of a given molarity. How would you make 100 mL of a 0.4 molar solution
of copper(II) sulfate? In this question, we notice something different. 0.40 M.
That italicized M not only stands for molarity, but it also stands for the units of molarity,
mol/L, and is pronounced “molar.” So, a 0.4 molar solution is the same as 0.4 mol/L The volume of the solution that we need to make is 100 mL,
which is the same as 0.1 liters. This problem is more complex than the previous
ones, but let’s go ahead and solve it like the
previous problems first. If we know V and M, then we can calculate n. M equals n over V rearranges to n equals MV. Substituting in for M and V give us 0.4 mol/L
times 0.1 liters. Liters on the top and bottom cancel, and the
answer is 0.04 moles. But, if we are going to make this in the lab,
we need to need to know how much copper(II) sulfate to weigh out,
so we need to convert the moles into grams. In order to do that, we need the molar mass,
which is our conversion factor. The easiest thing to do would be to look at
the chemical label, which almost always has the molar mass written on it; but let’s suppose we can’t do that. The Roman numeral II means that the copper
has a +2 charge, and sulfate is SO4 two minus. Because the charges are the same, the compound
is simply CuSO4. However, copper(II) sulfate is almost always
found as the pentahydrate, which means it has 5 water molecules that
have cocrystallized with the copper(II) sulfate. These water molecules aren’t really doing
anything; they are basically just hitchhiking with the
copper(II) sulfate. However, these water molecules do affect the
molar mass, which we will go ahead and calculate now. The formula contains one copper, with a molar
mass of 63.55 g/mol, one sulfur which is 32.07 g/mol,
9 oxygens, so we multiply 9 times 16 which equals 144 g/mol,
and 10 hydrogens, so we multiply 10 times 1.008 to get 10.08 g/mol. Add all of these together to get the molar
mass of 249.7 g/mol. This is our conversion factor between moles
and grams. In order to calculate the mass of copper(II)
sulfate needed we multiply 0.04 moles by a conversion factor
fraction. Because we are staring with moles,
one mole goes on the bottom. Because we are solving for grams, 249.7 grams
goes on the top. Moles on the top and bottom cancel out. Because the one is on the bottom, multiply
0.04 times 249.7 to get 9.988 grams. So how to you actually make the solution? First, get a 100 mL volumetric flask with
a stopper. It should look something like this. Next, weigh out 9.988 grams of copper(II)
sulfate pentahydrate, and carefully add all of that to the flask. You may need to use a funnel. Then, add distilled water to the flask until
the water reaches the fill line. Put the stopper on the flask,
and with your thumb on the stopper gently hold the flask upside down then right side
up. Repeat this a few times. Copper(II) sulfate tends to dissolve slowly
unless you have a fine powder, so you may want to just set it down and let
it dissolve on its own. Eventually, it will form a nice blue solution. Thanks for watching. If this video helped you at all,
please give me a thumbs up. Feel free to also share any comments or questions
you have below, subscribe to my channel,
or check me out at

100 thoughts on “Molarity Made Easy: How to Calculate Molarity and Make Solutions

  1. Good presentation that includes all the aspects that a student needs to know for a thorough understanding of molarity and concentration. I particularly like the variety of examples provided that demonstrated the different ways in which molarity could be use. the video is short but yet so enlightening. Good job!

  2. The video is useful, but the random appearance of a pentahydrate which was not part of the actual question kind of screws up everything. It's effectively adding a random integer.

  3. If we are to assume the 5H2O is present in CuSO4 then, and you said there were exception(s), under what circumstances is it assumed that the pentahydrate is not present? -Laboratory Science Major

  4. Great video, and easier to understand than my actual chemistry teacher haha. Though I realize that I am pretty late to jump on the bandwagon, I was wondering if you can help me with this question:
    What is the molarity of the KOH solution if 20.0mL neutralizes 7mL of 1.2 M H3PO3
    Sorry if this question is actually easier than I'm making it 😅 again, great video. Thank you!

  5. Excellent presentation very easy to understand sir thank you for your hard work t make easy to understand

  6. Taking Clinical Chemistry and had some review problems like this. Only problem is I took Chem 2 YEARS ago for my first degree. This video explained it so much better than my lecture. Thank you!

  7. Brother there is a question for you … I think you are well expert in chemistry ..there is very serious problem with tomato ketchup its become dual and black after 04 month due to oxidation and unsuitability of lycopin and carotinoide pigments ..can you please suggest me some thing better …that how can we sustain the colour of tomato ketchup for up to 07 month during storage

  8. Thank you for this video!!!
    I take Chemistry online, and the book is so confusing. I wish the book/online program was a clear as this. Doesn't make sense to pay nearly $200 for a digital book, then have to look to the Internet & Youtube for answers…

  9. How would you work out these problems?
    3) Consider a 0.82 M aqueous solution of glucose (C6H12O6).
    a) Convert this concentration to mM.
    b) Convert this concentration to g/L.
    c) Convert this concentration to ppm.
    d) Convert this concentration to % w/w.

  10. I watched this video three times over and finally understood the concept, thanks to Ketzbook and my lovely chemistry teacher. Everything makes so much more sense, and I passed my chem test, thanks Ketzbook!

  11. This is good, but we rarely start with how many moles of a solute are used to make a solution. Should start with the basics of converting grams to moles, first.

  12. Yeah I’m a sophomore in high school and I have a college chemistry final tomorrow. Pretty sure I’m going to fail, oh well 🤷‍♂️

  13. Your way of presentation is awesome. Making complicated topics into easier. Thanks.
    Expecting more topics from thermodynamics and complex compounds.

  14. the last answer is wrong!
    the number of moles will be 4 and not 0.04 at 5:43. So the mass is also incorrect right?
    However the video was awesome. Thank you for clearing my doubts sir!

  15. Thank you,
    I understand it much better than my teacher.

    I've a Que. – how to prepair a solution of 6 milimolar of AgNo3.

  16. Greetings🙌, Sir I really appreciated all yor hard work cos this is a spoon feeding to some of us. Furthermore i 've queri and I'm kind of confused on morality/concentration formula cos I came crossed one formula which is,

    Morality = # of moles÷ Volume × 1000.
    E.g. 0.050÷125×100=0.4dm^3. Is it a correct formula to use coz compare to yor answer you end with .0.40,


    FORMulA = moles÷ molarity × 1000 e.g. 0.12÷25×1000= 4.8, will I still get 3/3 without Ending with Zero?? And what ab2 De dm^3 ???????

  17. This is a very good lesson! A have one objection at the end: It is better to put the powder of CuSO4*5H2O to some partly filled with water 100 ml volumetric flask so that it doesn't stuck to the bottom and to have space to agitate the solution before making the volume.

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