Calculations Using Molarity
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Calculations Using Molarity

There are several types of calculations that you need to be able to do with molarity.

First, you should be able to calculate the molarity if you are given the components of the solution.
Second, you should be able to calculate the amount of solute in (or needed to make) a certain volume of solution.
Third, you might need to calculate the volume of a particular solution sample.
Fourth, you might need to calculate the concentration of a solution made by the dilution of another solution. This and related calculations will be covered in a separate page.

In either of the first two cases, the amount of solute might be in moles or grams and the amount of solution might be in liters or milliliters. Please note that with molarity we are concerned with how much solute there is and with how much solution there is, but not with how much solvent there is.

Examples (Ex. 3)

The examples that follow are also shown in example 3 in your workbook. You may want to look through the workbook examples on your own and if they make perfect sense to you, test your understanding by doing exercise 4. Then check your answers below before continuing on to dilution calculations.

General Relationship (Ex. 3a)

Here is the general relationship that you will be using over and over again. The molarity is equal to the number of moles of solute divided by the volume of the solution measured in liters. If you like to think of numbers and units instead of quantities look at the second version of the equation. In this equation x, y and z represent numbers: 2, 6 and 3 for example.

molarity =

moles of solute
liter of solution

x M =

y moles
    z L

2 M =

6 moles
    3 L

Calculating Molarity from Moles and Volume (Ex. 3b)

Here we are given something to figure out.  To get the molarity we need to divide the number of moles of NaCl by the volume of the solution. In this case that is 0.32 moles NaCl divided by 3.4 L, and that gives 0.094 M NaCl.
What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 liters?
molarity = 0.32 moles NaCl
        3.4 L
= 0.094 M NaCl

Calculating Molarity from Mass and Volume (Ex. 3c)

This one is a bit more difficult. To get molarity we still need to divide moles of solute by volume of solution. But this time we're not given the moles of solute. We have to calculate it from the mass of NaCl. We multiply 2.5 g NaCl by the conversion factor of 1 mole NaCl over the formula weight of NaCl, 58.5 g. That tells us that we have 0.0427 mole of NaCl. I kept an extra digit here because we are not done with the calculations. When we are done I'll round off to two digits, the same as in the 2.5 g weight of NaCl. Now that we know the moles we can calculate the molarity. Moles of solute (0.0427) divided by the volume of the solution (0.125 L) gives us 0.34 M NaCl.
What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water to make 125 ml of solution?

molarity =

moles of solute
liter of solution

2.5 g NaCl x

1 mole NaCl
58.5 g NaCl

= 0.0427 mole

molarity =

0.0427 mole NaCl
       0.125 L

=

0.34 M NaCl

Calculating Mass of Solute from Molarity (Ex. 3d)

This question asks how you would prepare 400. ml of 1.20 M solution of sodium chloride. In this case what you need to find out is how much NaCl would have to be dissolved in 400 ml to give the concentration that is specified. This amount is going to have to be in grams because we don't have any balances that weigh in moles. So there is more than one step to this problem.
 

The approach shown here is a conversion factor approach. It involves remembering that molarity is a relationship between moles and liters. 1.20 M NaCl means there is 1.2 moles of NaCl per 1.00 liter of solution. We can use that as a conversion factor to set up the calculation that relates 400. ml (or .400 L) to the appropriate number of moles of NaCl. So we take .400 L and multiply by the conversion factor to get .480 moles NaCl. The next step is to find out how many grams that is. We change from moles of NaCl to grams by using the formula weight. It comes out to 28.1 g NaCl. So the answer is that you would make the solution by dissolving 28.1 g NaCl in enough water to make 400 ml of solution.

 

How would you prepare 400. ml of 1.20 M solution of sodium chloride?

Remember:

1.20 M NaCl =

1.20 moles NaCl
1.00 L solution

0.400 L solution x

1.20 moles NaCl
  1.00 L solution

=

0.480 moles NaCl

0.480 moles NaCl x

58.5 g NaCl
1 mole NaCl

=

28.1 g NaCl
Dissolve 28.1 g NaCl in enough water to make 400 mL of solution.
There is also more than one way to do this problem. If you like the algebraic approach, you would write down the general equation shown in part a, substitute in the known values, solve for moles of NaCl, and then change that into grams.
Algebraic approach:
Set up general equation:

x M =

y moles
    z L
Substitute known values:

1.20 M =

y moles
0.400 L
Solve for moles:

y moles =

1.20 M x 0.400 L = 0.480 moles
Change to grams:

0.480 mole x 58.5 g/mole = 28.1 g

 

Calculating Moles of Solute from Molarity (Ex. 3e)

This question is a little easier. We do it the same way as the first step of the previous problem and then we stop. To find out how many moles of salt are contained in 300. ml of a 0.40 M NaCl solution, we start with the volume in liters (0.300 L) and multiply it by the number of moles per liter of solution, which is 0.40 moles over 1.00 L. The answer is 0.12 moles of NaCl. This could also have been done using algebra by writing down the general equation relating molarity, moles and liters, substituting the known values, and then solving the equation for moles.
How many moles of salt are contained
in 300. mL of a 0.40 M NaCl solution?

0.300 L x

0.40 moles NaCl
1.00 L solution

=

0.12 moles NaCl

Practice (Ex. 4)

Now you should take some time to review the calculations above (ex. 3). If you have any questions, check with your instructor. Once you are familiar with how those are done then you should try answering the following questions (exercise 4 in your workbook). Get help from the instructor if you need it. Check your answers below before you continue with the lesson.

Molarity Calculations: Practice

How would you prepare 100. mL of 0.25 M KNO3 solution?
A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution. What is the molarity of the solution?
How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution?
Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400. mL of this solution?

 

Answers (Ex. 4)

Here are the answers to exercise 4.

How would you prepare 100. mL of 0.25 M KNO3 solution?  Dissolve 2.53 g of KNO3 in enough water to make 100 ml of solution.
A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution. What is the molarity of the solution? 0.324 M
How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution? 
0.0308 mol
Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400. mL of this solution?  120 g

 

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