Solution Preparation and Dilutions
Part A. Making Solutions using Mass/Volume Concentrations
Many solutions are prepared with a certain mass of solute in a certain volume of
solvent. Any metric mass in any metric volume is possible, but the most common units
of mass/volume concentrations are as follows:
g/mL
g/L
mg/mL
μg/mL
μg/μL
ng/L
ng/μL
grams per milliliter
grams per liter
milligrams per milliliter
micrograms per milliliter
micrograms per microliter
nanograms per liter
nanograms per microliter
To determine how to prepare a certain volume of a solution at a certain mass/volume
concentration, use the equation below. Convert units as necessary to make sure units
that are used can be cancelled out.
Mass/Volume Concentration Equation
concentration desired x total volume desired = mass of solute in the total volume
desired
(ex. g/mL)
(ex. mL)
(ex. g)
Ex. A technician needs 50 mL of 15 mg/mL pepsin solution for an experiment. Using the
Mass/Volume Equation, the calculation would be as follows:
15 mg/mL x 50 mL = 750 mg = 0.75 g pepsin
The technician would add 0.75 g pepsin to a container and fill up to the 50 mL mark with
solvent (usually deionized water). Most balances weigh in grams, so the conversion
from mg to g was necessary.
Determine the calculations for the solutions in the Practice Problems section for Part A.
Making Solutions Using Mass/Volume Concentrations at the end of this handout.
Part B. Making SolutionsUsing Molarity Concentrations
The concentration of many solutions is reported as moles/liter (mol/L or M; the M is
spoken “molar”) or some function of those units. This concentration measurement is
called molarity. Molarity is sometimes a challenging concept to understand. However,
with your recently acquired solution preparation skills, you will see that making molar
solutions requires only one extra calculation.
Solution Preparation 1
To understand how to make a solution of a given molarity, you must know what a
“mole” is. A mole of a compound is equal to 6.02 x 1023 molecules, but that is not really
a very useful number. So, in biotechnology, it is easier to use this definition: The unit “1
mole” is the mass, in grams, equal to the molecular weight (MW), also called “formula
weight” (FW), of the substance. The FW can be determined by using a Periodic Table or
by adding the atomic weights of the atoms in the molecule. An easy way, though, is to
just read the label of a chemical reagent bottle, which lists the “MW” or “FW.” The
molecular weight of NaCl is 58.5 atomic mass units (amu) since the Na atom weighs 23
amu, and a Cl atom weighs 35.5 amu.
Molarity concentrations are reported as the number of moles per liter (mol/L or M). If
the concentration is very low, then the concentration could be reported in
millimoles/liter (mmol/L or mM). If you wanted a 1-M NaCl solution, you would
measure out 1 mole of NaCl (58.5 g) and dissolve it in water to a total volume of 1 L.
This gives you 1 mole of NaCl per liter of solution, 1 M NaCl.
A liter of solution is a large volume for most research and development purposes. In
research and development labs, mL or μL quantities are usually used. To determine
how to mix up a smaller volume of a solution of some molarity, follow the example
below.
Multiply the volume desired (L) by the concentration (molarity) desired (mol/L), as you
did in the mass volume calculations. Then, multiply the result by the compound’s
molecular weight (g/mol) to account for measuring in moles, as in the following
equation:
Molarity Concentration Equation
volume x molarity x molecular weight = grams of solute to be dissolved in
wanted
desired
of the solute
solvent to the final desired volume
(L)
(mol/L)
(g/mol)
Convert smaller or larger units to these as necessary. The “L” units cancel out and the
“mol” units cancel out, leaving the mass in grams of the solutes needed to make the
solution.
Ex. A technician needs 50 mL of 0.5 M NaCl solution for an experiment. Using the
Molarity Concentration Equation, the calculation would be as follows:
0.05 L x 0.5 mol/L x 58.5 g/mol = 1.46 g NaCl
The technician would add 1.46 g NaCl to a container and fill up to the 50 mL mark with
solvent.
Determine the calculations for the solutions in the Practice Problems section for Part B.
Making Solutions Using Molarity Concentrations at the end of this handout.
Solution Preparation 2
Part C. Making Dilutions of Concentrated Solutions
Making dilutions of concentrated solutions is a common practice in a biotechnology lab.
A concentrated solution is generally called a “stock solution,” and the diluted solution is
called the “working solution.” Preparing a concentrated stock solution saves a lot of
time and is easier to store than large volumes of diluted working solutions. Making a
working solution simply requires diluting some volume of stock solution to the
concentration needed.
When a number of dilutions must be made, and each is proportionally the same dilution
as the one before, it is called a serial dilution. Doing a serial dilution makes sense for
many experiments when many samples of varying concentrations are needed. A serial
dilution is also useful for preparing very dilute solutions that are hard to make from
scratch, because the solute masses can be too small to measure on a balance.
Each succeeding sample is made with the
same ratio of sample and diluent as the one before
500 mL
1M NaCl
1 M NaCl
500 mL
0.5M NaCl
0.5 M NaCl
500 mL
0.25M NaCl
0.25 M NaCl
0.125 M NaCl
Each of the above dilutions is one part previous sample and one part solvent. This is
called a 1:2 dilution, or one part sample to two total parts.
To figure out how to prepare a working solution from a stock solution, we use the
process bulleted below:
 Restate the problem:
o What do you have, i.e., what is the concentration of the stock solution and
how much of it do you have?
o What is needed, i.e., what is the concentration of the working solution and
how much of it do you need?
 Convert all concentrations and volumes to the same units.
 Calculate the Dilution Factor: Concentration of stock/Concentration of working
solution
 Calculate the volume of stock needed: Volume needed/Dilution factor
 Calculate the volume of solvent (usually water) needed: Volume of working
solution needed – Volume of stock needed
For example, a technician needs to prepare 150 mL of 0.1 M TRIS (the working solution),
from 100 ml of a stock soltuion of 1 M TRIS.
 What you have: 100 ml of 1.0 M Tris
 What you need: 150 ml of 0.1M Tris
 All volumes and concentrations are already in the same units
Solution Preparation 3



The dilution factor = 1M Tris/0.1M Tris = 10
The volume of stock needed = 100/10 = 10ml
The volume of solvent needed = 100 ml – 10 ml = 90 ml of water.
Determine the calculations for the solutions in the Practice Problems section for Part C.
Making Dilutions of Concentrated Solutions at the end of this handout.
Solution Preparation 4
Practice Problems
Part A. Making Solutions Using Mass/Volume Concentrations
1. Describe how you would prepare 25 mL of a NaCl solution at a concentration of 2.5
g/mL.
2. Describe how you would prepare 2 L of a 0.5 g/mL dextrose solution.
Part B. Making Solutions Using Molarity Concentrations
1. Describe how you would prepare 125 mL of a 10 M NaOH solution.
2. Describe how you would prepare 75 mL of a 0.1 M NaCl solution.
Solution Preparation 5
Part C. Making Dilutions of Concentrated Solutions
1. Describe how you would prepare 950 mL of a 1M CuSO4•5H2O solution from a 25M
CuSO4•5H2O stock.
2. Describe how you would prepare 50 mL of a 5 mM NaCl solution from a 1 M NaCl
stock.
Solution Preparation 6