Solutions and Dilutions
Solution Terminology:
Solute: That compound that is to be made into a solution. Can be solid (usually) or liquid.Solvent: That which the
solute is dissolved in; usually water but can be buffers or organic solvents.
Solution: A known mixture of solute and solvent.
Molar Solutions: A mole of a compound dissolved in enough water or other solvent to make exactly one liter (L; 1000
mL) of solution at 20 C is a molar (M) solution.
Percentage Solutions: X % = X g of solute in 100 mL of solution. If solute is only available as a liquid, X % = X mL of
solute in 100 mL of solution (simply subtract the volume of solute from 100 mL and add that volume of solvent). If
solid solute is not a large amount, safe to just add 100 mL of solvent and assume total volume is 100 mL.
Other Solutions: Because their molecular weights are so large or unknown, Macromolecules such as DNA, RNA,
polysaccharides, lipids, or proteins are usually expressed in terms of milligram (mg) or microgram (g) per mL or
nanogram (ng) solution or, if an enzyme, units of activity per mL.
Low concentrations: Parts per thousand (ppt), parts per million (ppm), parts per billion (ppb) are convenient units for
expression concentrations especially for extremely dilute solutions, for example, in expressing the concentrations of
trace chemicals, particularly pollutants and toxic chemicals in solutions and also air.
Solutions: We will review how to make solutions and determine their concentrations: molar solution (M, mM, M, nM)
, % solutions (weight to volume; volume to volume), weight solutions (mg/mL, g/mL, ng/mL), ppt, ppm, ppb. Use
appropriate scientific notation and significant figures in the below calculations and answers.
A. Molar Solutions: A molar (M) solution is define as a mole of a compound dissolved in enough water to make
exactly one liter of solution at 200 C. A mole is the molecular weight of a compound in grams (g), e.g. for
sodium chloride, it is 22.99 g (atomic weight of sodium) + 35.46 g (atomic weight of chlorine) = 58.45 g.
Dissolve 58.45 g of NaCl in 1 liter (1 L) of solution = 1 molar (M) solution. You could also make up the same
molarity of NaCl by dissolving 5.845 g NaCl in 100 milliliter (mL) or 0.5845 g in 10 mL or 0.05845 g in 1
mL. To make up lesser concentrations, say 0.1 M, you would add 5.845 g NaCl in 1 liter (L) solution or 0.5845
in 100 mL or 0.05845 g in 10 mL or 0.005845 g. in 1 mL. The metric sytstem is also useful in solutions too: 1
molar (M) is the same as 1,000 millimolar (mM) or 1,000,000 micromolar (M) or 1,000,000,000 nanomolar
(nM). Answer the following questions on a separate piece of paper for your lab report. Use correct scientific
notation and significant figures:
6. Exercise: Explain how you would make 1000 mL of a 0.25 M solution containing sucrose (C12H22O11 ).
7: Explain how you would make up 200 mL of 0.25 M solution containing glucose using solid sucrose and
distilled water,
8. Explain how you would make up 200 mL of 0.025 M solution containing glucose using solid sucrose and
distilled water.
9. What is the concentration in 8 expressed as mM?
10. What is the concentration in 8 expressed as M?
11. What is the concentration in 8 expressed as nM?
B. Percentage Solutions: based on volume of solution. A 12.5% solution of sucrose is 12.5 g of sucrose in a total
volume of 100 mL. Formula to remember: X% = X gram (g) per 100 mL of solution. Use correct scientific
notation and significant figures:
12. (Worth 4).What is the molarity (M) of 12.5% solution of sucrose? mM? M? nM?
D. Other Solutions: Because their molecular weights are so large or sometimes unknown, Macromolecules such as
polysaccharides, DNA, RNA, and proteins are usually expressed in terms of mg or microgram (g) per mL
solution or, if an enzyme, units of activity per mL. Use correct scientific notation and significant figures:
13. Explain how you would make 20.0 mL of a solution that contains 4.0 mg of cholesterol per mL of ethanol.
14. Using scientific notation and significant figures: What is the concentration in #13 in g per mL?
Concentrated Solutions and Dilutions
The concept of concentrated solutions and then the dilution of these concentrated solutions to give working solution can
be confusing (Just ask my previous Microbiology, Cell Biology, Biochemistry, and Honors and Undergraduate
Research students!). Concentrated or stock solutions are very useful in biology since it allows you to make up a
bulk solution that is used often in a much smaller volume, and then use these stock solutions to prepare working
solutions as needed. What follows is one technique to calculate the dilution ratio of concentrated or stock solutions
to give a working solution based on a very simple formula. There are certainly other ways to do this, and you may
have your own way, but this one may be the most helpful one for you at this stage.
Dilution terminology:
Stock solution: concentrated solution which is to be diluted to give either the secondary solution or working solution;
can be expressed in M or % or even ?X concentrate where ? is some number greater than 1, e.g., 10X means that the
stock solution is 10 times concentrated than what you ordinarily work at 1X. The amounts of each component in the
1X solution are usually written in your lab book and not on the bottle.
Secondary solution: an intermediate solution of the stock solution, to be used to make the working solution. Usually
still some X fold more concentrated than you want, e.g. 2X. A secondary solution is usually made if you determine
that you do not have the accurate pipetting equipment to dilute the Stock solution directly to the volume of working
solution needed.
Working solution: diluted solution that you want, to be made from the concentrated solution, ready to use. Usually at 1
X concentrate but could be less, e.g., 0.5X depending on what the experiment requires.
Diluent: solution used for diluting Stock solution to make the secondary and working solutions, usually water but could
be a buffer of some sort or even organic solvents.
Serial dilutions: a sequential set of dilutions in which the stock for each succeeding dilution is the working solution from
the previous one. In effect, except for the last dilution, each tube is both a stock and a working solution. For
example, 2-fold dilutions means that you start with the original solution and dilute it several times using the same
dilution. A 2-fold dilution is one part or mL of the stock solution with one part or 1 mL of diluent.
For Dilutions Only: Using the algebraic formula below, one can calculate the volume, V1, of the initial stock solution
or C1 to give the final working or secondary stock concentration, C2, and final working or secondary stock volume, V2.
The Difference of V2 and V1 will give you the volume of water to add to V1 to give the total volume V2. Note that the
Vs and Cs have to be in the same units in order to get the correct answers. Put your units down and you should end up
with the unit that is asked for!
C1 X V1 = C2 X V2
where C1 = the initial concentration (M, mM, M, nM%, mg/mL, ?X concentrate), V1 = volume of initial stock C1
required (mL, L, L, nL), C2 = final concentration required (mL, L, L, nL), V2 = final volume needed of C2
(ditto) and X is the multiplication sign. Note that C1 and C2 have to be in the same units and V1 and V2 have to be
in the same units.
You can also use simple algebra relationship to solve for one unknown. For example, if you have a solution of 0.5 M
solution, what is the concentration of the solution in mM? There is 1000 mM in 1 M. Thus
X/0.5 M = 1000 mM/1 M, solve for X!
X = ? mM
Notice that the units will cancel and you end up with the number and the correct units.
Always put your units down for all known numbers and you can’t go wrong in your calculations!
Examples (try these but we will work on some of them during the next laboratory); include these answers in your lab
reports on separate sheets of paper. Use correct scientific notation and significant figures.
41. Given a stock concentration of 0.500 M NaCl, how would you make 700.0 mL of a working solution of 40 mM
NaCl.
42. A stock solution of 10.00X TAE buffer (used in molecular biology) is used to make 1.00 X TAE buffer. How would
you prepare 800.0 mL of 1X TAE buffer working solution from a 10.00X TAE stock?
43. How much of a 10.00 mg/mL stock solution of ampicillin would you need to add to 1000.0 mL of bacterial growth
medium to achieve a final concentration of 25.0 microgram/mL ampicillin as a working solution?
44. You have a small amount of a stock bovine serum albumin (BSA) solution at a concentration of 5.0 mg/mL. You
want 6.00 mL of BSA at a concentration of 10.00 ng/mL. Describe how you would serially dilute your sample to
reach 10.00 ng/mL in your working solution. The smallest amount that you can pipette accurately is 1 L! Hint:
you have to make a secondary stock solution to solve this problem/
45. How would you make 200.0 mL of a working solution of ethanol at 60.0 % from a stock solution of 80.0 %
ethanol?
46. Describe how you would prepared a 500.0 mL of 10.00X stock solution in which the working solution is 1.00 X and
is 1.50% NaCL.
47. If a 20.0 X stock solution is 0.200 M, what is the final concentration of a 1.00 X working solution?
48. Given a stock solution of 14.00% NaCl, how would you make 400.0 mL of 5.00 % NaCl working solution?