Hi there, I was just wondering shouldnt the answer in example 1 be 0.48 mol/Litre and NOT 4.8 M? I get the same answer to the last step before the answer, but when i do the calculation i get 0.48 mol/litre. thank you so much.

I believe you're correct. There was likely a typographic error in the example. We see in the previous step the conversion was done correctly (50 mL = .050 L) so we have 0.02401 mol / .050 L. A quick check with the calculator shows that this is in fact 0.48 mol/L or 0.48 M.

I was told in school that molarity should be moles/dm^3, but is this different from moles/litres?

A liter is equal to a cubic decimeter, so it is the same.

So this isn't quite the right place for my question, but I can't find the right place for the life of me... If someone could maybe point me to a video/article on converting between concentration units, especially molarity to ppt or ppm, that'd be great. In the mean time, I've been asked to take a known molarity of a solution and convert it into parts per thousand. I tried Google and I /think/ I got the right formula but I'm not positive, so can someone check it for me please? So what I did was start with my given molarity as mol/L. I assumed there wouldn't be enough solute to drastically affect density and so I changed 1 L to 1000g, so I now have mol/1000g. Then I multiply by the molar mass of the solute (NaOH - 39.998) so I'm now g NaOH/1000g solution. Then I multiply the whole thing by 1000 to get ppt, right? Sort of like calculating a percent?

You did it almost perfectly. A concentration of 1 g NaOH/1000 g solution is 1 g per 1000 g or one part per thousand (1 ppt) — no need to multiply by 1000. In the same way, a concentration of 1 g per 100 g is one part per hundred (1 %).

in hint one how do you know there is .1L of solute?

There must have been a typo. I think in the description they meant 0.100L instead of 0.100mL.

How would you find the molarity of SO2 if you have it dissolved in 100 grams of water at 85 degrees Celcius?

Assuming that you do not know the amount of SO2 that was dissolved to prepare the solution, you may try to invoke Henry's Law and determine the concentration of SO2 in the headspace (just above) of the solution. For a primer on Henry's Law, you can check out this article: https://www.thoughtco.com/henrys-law-example-problem-609500 You can also check these links below for sample procedures on determining the amount of SO2 vapor (<- what causes acid rain!): https://www.law.cornell.edu/cfr/text/40/appendix-A-2_to_part_50 https://www.astm.org/Standards/D2914.htm Finally, you can check this link, so you can convert your determined SO2 vapor concentration to SO2 molarity in water: https://www.ems.psu.edu/~brune/m532/m532_ch5_aqueous_phase.htm Yeah, this is some detective work (and a lot of hard work!).

Question: Is this just coincidence, or does this make sense... In the equation, we have 1 Pb(NO3)2 + 2 KI...we have twice as many KI as Pb(NO3)2. Since we have 0.1L of 1Pb(NO3)2, can I just multiply the 0.1 L x 2, since we use twice as much KI as we do Pb(NO3)2? (Or if the equation happened to have 4KI, could we simply multiply 0.1L x 4)? Thanks for the help!

What you suggest is fine just as long as the concentrations of the two solutions are the same. But if, say, the Pb(NO3)2 solution was twice the strength of the KI solution then you would only need 0.1 L of each to get the same number of moles.

I understood what molarity is quite well......but what is normality, formality and molarity? If we have molarity why are they even needed then?

With any luck, like most people, you will be able to safely ignore normality and formality. Molality is moles / mass of solvent (SI unit: mol/kg) -- for use see: https://en.wikipedia.org/wiki/Molality#Usage_considerations Normality is explained here: https://en.wikipedia.org/w/index.php?title=Equivalent_concentration&redirect=no Formality is more or less totally ignored and often when we say molarity we actually mean formality see: http://www.chemiasoft.com/chemd/node/25 A good discussion of most of these is here: https://socratic.org/questions/what-is-molarity-molality-and-normality

What is the difference between molarity and molality?

Molarity is (mol of solute)/(L of solution). Molality is (mol of solute)/(kg of solvent). More on the difference here: https://www.khanacademy.org/science/health-and-medicine/lab-values/v/molarity-vs-molality

Main content

Course: Chemistry library > Unit 9

Lesson 3: Mixtures and solutions

Molarity

Q: Question: Is this just coincidence, or does this make sense... In the equation, we have 1 Pb(NO3)2 + 2 KI...we have twice as many KI as Pb(NO3)2. Since we have 0.1L of 1Pb(NO3)2, can I just multiply the 0.1 L x 2, since we use twice as much KI as we do Pb(NO3)2? (Or if the equation happened to have 4KI, could we simply multiply 0.1L x 4)? Thanks for the help!

What you suggest is fine just as long as the concentrations of the two solutions are the same. But if, say, the Pb(NO3)2 solution was twice the strength of the KI solution then you would only need 0.1 L of each to get the same number of moles.

Q: I understood what molarity is quite well......but what is normality, formality and molarity? If we have molarity why are they even needed then?

With any luck, like most people, you will be able to safely ignore normality and formality. Molality is moles / mass of solvent (SI unit: mol/kg) -- for use see: https://en.wikipedia.org/wiki/Molality#Usage_considerations Normality is explained here: https://en.wikipedia.org/w/index.php?title=Equivalent_concentration&redirect=no Formality is more or less totally ignored and often when we say molarity we actually mean formality see: http://www.chemiasoft.com/chemd/node/25 A good discussion of most of these is here: https://socratic.org/questions/what-is-molarity-molality-and-normality

Q: What is the difference between molarity and molality?

Molarity is (mol of solute)/(L of solution). Molality is (mol of solute)/(kg of solvent). More on the difference here: https://www.khanacademy.org/science/health-and-medicine/lab-values/v/molarity-vs-molality

Google Classroom

Definitions of solution, solute, and solvent. How molarity is used to quantify the concentration of solute, and how to calculate molarity.

Key points

Mixtures with uniform composition are called homogeneous mixtures or solutions.
Mixtures with non-uniform composition are heterogeneous mixtures.
The chemical in the mixture that is present in the largest amount is called the solvent, and the other components are called solutes.
Molarity or molar concentration is the number of moles of solute per liter of solution, which can be calculated using the following equation:

$Molarity = \frac{mol solute}{L of solution}$ ‍

Molar concentration can be used to convert between the mass or moles of solute and the volume of the solution.

Introduction: Mixtures and solutions

In real life, we often encounter substances that are mixtures of different elements and compounds. One example of a mixture is the human body. Did you know that the human body is approximately

57 %

water by mass? We are basically an assortment of biological molecules, gases, and inorganic ions dissolved in water. I don't know about you, but I find that pretty mind-boggling!

If substances are mixed together in such a way that the composition is the same throughout the sample, they are called homogeneous mixtures. In contrast, a mixture that does not have a uniform composition throughout the sample is called heterogeneous.

Homogeneous mixtures are also known as solutions, and solutions can contain components that are solids, liquids and/or gases. We often want to be able to quantify the amount of a species that is in the solution, which is called the concentration of that species. In this article, we'll look at how to describe solutions quantitatively, and discuss how that information can be used when doing stoichiometric calculations.

Molar concentration

Nope!

Molar concentration is the same as molarity, but molarity and molality are not the same thing. They are different ways to quantify the amount of solute in a solution, and the concentration of a solution in molarity is not interchangeable with its concentration in molality. In this article we are only discussing molarity.

For more on the difference between the two definitions, see this video on molarity vs. molality.

The component of a solution that is present in the largest amount is known as the solvent. Any chemical species mixed in the solvent is called a solute, and solutes can be gases, liquids, or solids. For example, Earth's atmosphere is a mixture of

78 %

nitrogen gas,

21 %

oxygen gas, and

1 %

argon, carbon dioxide, and other gases. We can think of the atmosphere as a solution where nitrogen gas is the solvent, and the solutes are oxygen, argon and carbon dioxide.

The molarity or molar concentration of a solute is defined as the number of moles of solute per liter of solution (not per liter of solvent!):

$Molarity = \frac{mol solute}{L of solution}$ ‍

You might think that it is because, well, we didn't take into account the volume of the solute! But actually, even if we added the volume of the solute to the volume of the solvent, we still might end up with a total volume that is different from the volume of the solution.

This is because the volume of a mixture depends on the intermolecular forces between the molecules in the mixture. This is especially true for a liquid solution, where the molecules are closer together and thus have strong interactions than in a solution of gases. Intermolecular forces between the solvent molecules are likely to be different than the intermolecular forces between the solvent and solute molecules. Depending on the nature of these interactions, the volume of the solution can be more or less than the volume of the solvent plus the volume of the solute!

Pretty crazy, but that's chemistry for you.

Molarity has units of

\frac{mol}{liter}

, which can be abbreviated as molar or

M

(pronounced "molar"). The molar concentration of the solute is sometimes abbreviated by putting square brackets around the chemical formula of the solute. For example, the concentration of chloride ions in a solution can be written as

[{Cl}^{-}]

. Molar concentration allows us to convert between the volume of the solution and the moles (or mass) of the solute.

Concept check: Bronze is an alloy that can be thought of as a solid solution of ~ $88 %$ ‍ copper mixed with $12 %$ ‍ tin. What is the solute and solvent in bronze?

Example 1: Calculating the molar concentration of a solute

Let's consider a solution made by dissolving

2.355 g

of sulfuric acid,

H_{2} {SO}_{4}

, in water. The total volume of the solution is

50.0 mL

. What is the molar concentration of sulfuric acid,

[H_{2} {SO}_{4}]

To find

[H_{2} {SO}_{4}]

we need to find out how many moles of sulfuric acid are in solution. We can convert the mass of the solute to moles using the molecular weight of sulfuric acid,

98.08 \frac{g}{mol}

${mol H}_{2} {SO}_{4} = 2.355 g H_{2} {SO}_{4} \times \frac{1 mol}{98.08 g} = 0.02401 {mol H}_{2} {SO}_{4}$ ‍

We can now plug in the moles of sulfuric acid and total volume of solution in the molarity equation to calculate the molar concentration of sulfuric acid:

$\begin{aligned} [H_{2} {SO}_{4}] & = \frac{mol solute}{L of solution} \\ = \frac{0.02401 mol}{0.050 L} \\ = 0.48 M \end{aligned}$ ‍

Concept check: What is the molar concentration of $H^{+}$ ‍ ions in a $4.8 {M H}_{2} {SO}_{4}$ ‍ solution?

Answering this question requires thinking about what is happening in solution. Since sulfuric acid is a strong acid, it dissociates completely into its constituent ions in solution,

H^{+} (a q)

and

{SO}_{4}^{-} (a q)

. This can also be written as follows:

H_{2} {SO}_{4} (a q) \to 2 H^{+} (a q) + {SO}_{4}^{2 -} (a q)

From this balanced equation, we can see for every one mole of sulfuric acid in solution, we end up with

2

moles of

H^{+} (a q)

. Using this ratio, the molar concentration of

H^{+} (a q)

will be:

[H^{+} (a q)] = 4.8 {M H}_{2} {SO}_{4} \times \frac{2 {mol H}^{+}}{1 {mol H}_{2} {SO}_{4}} = 9.6 {M H}^{+} (a q)

Therefore, our molar concentration of

H^{+} (a q)

is twice the molar concentration of

H_{2} {SO}_{4}

Example 2: Making a solution with a specific concentration

Sometimes we have a desired concentration and volume of solution, and we want to know how much solute we need to make the solution. In that case, we can rearrange the molarity equation to solve for the moles of solute.

$mol solute = Molarity \times L of solution$ ‍

For example, let's say we want to make

0.250 L

of an aqueous solution with

[NaCl] = 0.800 M

. What mass of the solute,

NaCl

, would we need to make this solution?

We can use the rearranged molarity equation to calculate the moles of

NaCl

needed for the specified concentration and volume:

$\begin{aligned} mol NaCl & = [NaCl] \times L of solution \\ = 0.800 \frac{mol}{L} \times 0.250 L \\ = 0.200 mol NaCl \end{aligned}$ ‍

We can then use the molecular weight of sodium chloride,

58.44 \frac{g}{mol}

, to convert from moles to grams of

NaCl

$Mass of NaCl = 0.200 mol \times \frac{58.44 g}{1 mol} = 11.7 g NaCl$ ‍

In practice, we could use this information to make our solution as follows:

Step

1.

Weigh out

11.7 g

of sodium chloride.

Step

2.

Transfer the sodium chloride to a clean, dry flask.

Step

3.

Add water to the

NaCl

until the total volume of the solution is

250 mL

Step

4.

Stir until the

NaCl

is completely dissolved.

The accuracy of our molar concentration depends on our choice of glassware, as well as the accuracy of the balance we use to measure out the solute. The glassware determines the accuracy of our solution volume. If we aren't being too picky, we might mix the solution in a Erlenmeyer flask or beaker. If we want to extremely precise, such as when making a standard solution for an analytical chemistry experiment, we would probably mix the solute and solvent in a volumetric flask (see picture below).

Summary

Mixtures with uniform composition are called homogeneous solutions.
Mixtures with non-uniform composition are heterogeneous mixtures.
The chemical in the mixture that is present in the largest amount is called the solvent, and the other components are called solutes.
Molarity or molar concentration is the number of moles of solute per liter of solution, which can be calculated using the following equation:

$Molarity = \frac{mol solute}{L of solution}$ ‍

Molar concentration can be used to convert between the mass or moles of solute and the volume of the solution.

Try it: The stoichiometry of a precipitation reaction

Molarity is a useful concept for stoichiometric calculations involving reactions in solution, such precipitation and neutralization reactions. For example, consider the precipitation reaction that occurs between

{Pb(NO}_{3})_{2} (a q)

and

KI (a q)

. When these two solutions are combined, bright yellow

{PbI}_{2} (s)

precipitates out of solution. The balanced equation for this reaction is:

${Pb(NO}_{3})_{2} (a q) + 2 KI (a q) \to {PbI}_{2} (s) + 2 {KNO}_{3} (a q)$ ‍

If we have $0.1 L$ ‍ of $0.10 {M Pb(NO}_{3})_{2}$ ‍, what volume of $0.10 M KI (a q)$ ‍ should we add to react with all the ${Pb(NO}_{3})_{2} (a q)$ ‍?

To determine what volume of

KI

solution we need, we can rearrange the molarity equation to solve for the total volume of solution:

$mol solute = Molarity \times L of solution$ ‍

If we divide both sides by the molarity of the solution, we get:

$L of solution = \frac{mol solute}{Molarity}$ ‍

We can plug in our known values for the moles of solute and the molarity:

$\begin{aligned} L of solution & = \frac{mol solute}{Molarity} \\ = \frac{0.020 mol}{0.10 M KI} \\ = 0.20 L KI \end{aligned}$ ‍

Thus, we need

0.20 L

of the

0.10 M KI

solution to completely react with

100 mL

0.10 {M Pb(NO}_{3})_{2}

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johannmlmstn
Posted 8 years ago. Direct link to johannmlmstn's post “Hi there, I was just wo...”
Hi there,
I was just wondering shouldnt the answer in example 1 be 0.48 mol/Litre
and NOT 4.8 M? I get the same answer to the last step before the answer, but when i do the calculation i get 0.48 mol/litre.
thank you so much.
Button navigates to signup pageComment on johannmlmstn's post “Hi there, I was just wo...”
(54 votes)
Answer
- Daniel Stoken
  Posted 8 years ago. Direct link to Daniel Stoken's post “I believe you're correct....”
  I believe you're correct. There was likely a typographic error in the example. We see in the previous step the conversion was done correctly (50 mL = .050 L) so we have 0.02401 mol / .050 L. A quick check with the calculator shows that this is in fact 0.48 mol/L or 0.48 M.
  Button navigates to signup page
  (33 votes)
Anson Chan
Posted 8 years ago. Direct link to Anson Chan's post “I was told in school that...”
I was told in school that molarity should be moles/dm^3, but is this different from moles/litres?
Button navigates to signup pageComment on Anson Chan's post “I was told in school that...”
(5 votes)
Answer
- Esther Dickey
  Posted 8 years ago. Direct link to Esther Dickey's post “A liter is equal to a cub...”
  A liter is equal to a cubic decimeter, so it is the same.
  Button navigates to signup page
  (31 votes)
Dawen
Posted 8 years ago. Direct link to Dawen's post “So this isn't quite the r...”
So this isn't quite the right place for my question, but I can't find the right place for the life of me... If someone could maybe point me to a video/article on converting between concentration units, especially molarity to ppt or ppm, that'd be great.

In the mean time, I've been asked to take a known molarity of a solution and convert it into parts per thousand. I tried Google and I /think/ I got the right formula but I'm not positive, so can someone check it for me please?

So what I did was start with my given molarity as mol/L. I assumed there wouldn't be enough solute to drastically affect density and so I changed 1 L to 1000g, so I now have mol/1000g. Then I multiply by the molar mass of the solute (NaOH - 39.998) so I'm now g NaOH/1000g solution. Then I multiply the whole thing by 1000 to get ppt, right? Sort of like calculating a percent?
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(5 votes)
Answer
- Ernest Zinck
  Posted 8 years ago. Direct link to Ernest Zinck's post “You did it almost perfect...”
  You did it almost perfectly.
  A concentration of 1 g NaOH/1000 g solution is 1 g per 1000 g or one part per thousand (1 ppt) — no need to multiply by 1000.
  In the same way, a concentration of 1 g per 100 g is one part per hundred (1 %).
  Comment on Ernest Zinck's post “You did it almost perfect...”
  (8 votes)
Rachel Silverman
Posted 8 years ago. Direct link to Rachel Silverman's post “in hint one how do you kn...”
in hint one how do you know there is .1L of solute?
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(4 votes)
Answer
- Philomath
  Posted 8 years ago. Direct link to Philomath's post “There must have been a ty...”
  There must have been a typo. I think in the description they meant 0.100L instead of 0.100mL.
  Button navigates to signup page
  (4 votes)
Abigail Baricevich
Posted 8 years ago. Direct link to Abigail Baricevich's post “How would you find the mo...”
How would you find the molarity of SO2 if you have it dissolved in 100 grams of water at 85 degrees Celcius?
Button navigates to signup pageComment on Abigail Baricevich's post “How would you find the mo...”
(2 votes)
Answer
- Hazelle R. Dela Cruz
  Posted 6 years ago. Direct link to Hazelle R. Dela Cruz's post “Assuming that you do not ...”
  Assuming that you do not know the amount of SO2 that was dissolved to prepare the solution, you may try to invoke Henry's Law and determine the concentration of SO2 in the headspace (just above) of the solution.
  
  For a primer on Henry's Law, you can check out this article:
  https://www.thoughtco.com/henrys-law-example-problem-609500
  
  You can also check these links below for sample procedures on determining the amount of SO2 vapor (<- what causes acid rain!):
  https://www.law.cornell.edu/cfr/text/40/appendix-A-2_to_part_50
  https://www.astm.org/Standards/D2914.htm
  
  Finally, you can check this link, so you can convert your determined SO2 vapor concentration to SO2 molarity in water:
  https://www.ems.psu.edu/~brune/m532/m532_ch5_aqueous_phase.htm
  
  Yeah, this is some detective work (and a lot of hard work!).
  Button navigates to signup page
  (2 votes)
Jeff Sellers
Posted 7 years ago. Direct link to Jeff Sellers's post “Question: Is this just c...”
Question: Is this just coincidence, or does this make sense...
In the equation, we have 1 Pb(NO3)2 + 2 KI...we have twice as many KI as Pb(NO3)2. Since we have 0.1L of 1Pb(NO3)2, can I just multiply the 0.1 L x 2, since we use twice as much KI as we do Pb(NO3)2? (Or if the equation happened to have 4KI, could we simply multiply 0.1L x 4)? Thanks for the help!
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(2 votes)
Answer
- RogerP
  Posted 7 years ago. Direct link to RogerP's post “What you suggest is fine ...”
  What you suggest is fine just as long as the concentrations of the two solutions are the same. But if, say, the Pb(NO3)2 solution was twice the strength of the KI solution then you would only need 0.1 L of each to get the same number of moles.
  Comment on RogerP's post “What you suggest is fine ...”
  (2 votes)
FoxFace
Posted 8 years ago. Direct link to FoxFace's post “I understood what molarit...”
I understood what molarity is quite well......but what is normality, formality and molarity? If we have molarity why are they even needed then?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- tyersome
  Posted 7 years ago. Direct link to tyersome's post “With any luck, like most ...”
  With any luck, like most people, you will be able to safely ignore normality and formality.
  
  Molality is moles / mass of solvent (SI unit: mol/kg) -- for use see: https://en.wikipedia.org/wiki/Molality#Usage_considerations
  
  Normality is explained here: https://en.wikipedia.org/w/index.php?title=Equivalent_concentration&redirect=no
  
  Formality is more or less totally ignored and often when we say molarity we actually mean formality see: http://www.chemiasoft.com/chemd/node/25
  
  A good discussion of most of these is here: https://socratic.org/questions/what-is-molarity-molality-and-normality
  Button navigates to signup page
  (2 votes)
Sevillano, Aida
Posted 4 years ago. Direct link to Sevillano, Aida's post “how do you find the volum...”
how do you find the volume when given the mass and M value
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(1 vote)
Answer
- Astic
  Posted 3 years ago. Direct link to Astic's post “We know that the formula ...”
  We know that the formula to calculate the molarity of a substance is M = n/V (n = moles, and V = volume of the solution).
  Rearranging the formula to make 'V' the subject allows us to figure out that V = n/M.
  When given the mass in Analytical Chemistry, we should always seek to covert the mass (given in any units) first into grams (if it is, then do not worry about this). We should then convert these grams into moles, to do so we require the molar mass of the solute, and dividing the given mass (in grams) by the molar mass provides us with the moles of the substance.
  Therefore, we have everything we need, we have calculated the moles, and we are already given the Molarity (M). Seek to substitute these values into their respective position within the rearranged equation above- V = n/M, calculating this value will output the volume.
  Button navigates to signup page
  (1 vote)
Artemis
Posted 8 years ago. Direct link to Artemis's post “What is the difference be...”
What is the difference between molarity and molality?
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(1 vote)
Answer
- Philomath
  Posted 8 years ago. Direct link to Philomath's post “Molarity is (mol of solut...”
  Molarity is (mol of solute)/(L of solution).
  Molality is (mol of solute)/(kg of solvent).
  More on the difference here: https://www.khanacademy.org/science/health-and-medicine/lab-values/v/molarity-vs-molality
  Button navigates to signup page
  (4 votes)
miARNr
Posted 6 years ago. Direct link to miARNr's post “Question1 :In a solution ...”
Question1 :In a solution with 2 species "A" and "B" ,with "A" having a greater number of moles but the "B" having a bigger molecular mass in such a way that it exceeds the mass of "A", who is the solvent ?
Question 2 : when 2 species are in the same amount , what determines who is the solvent ?
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(2 votes)
Answer
- cali24
  Posted 4 years ago. Direct link to cali24's post “For Question 2, I believe...”
  For Question 2, I believe that the substance you are using as the base is the solvent. For example, if you have 50 g of water and 50 g of salt, then the solvent would be the water, as you put the salt IN the water, not the water IN the salt.
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  (1 vote)

Course: Chemistry library > Unit 9

Molarity

Key points

Introduction: Mixtures and solutions

Molar concentration

Example 1: Calculating the molar concentration of a solute

Example 2: Making a solution with a specific concentration

Summary

Attributions

Other references

Try it: The stoichiometry of a precipitation reaction

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