Measure |
Symbol |
Relative Length |
Exponential Notation |
Meter |
M |
1 |
10 |
Decimeter |
dm |
.1 |
10 |
Centimeter |
cm |
.01 |
10 |
Millimeter |
mm |
.001 |
10 |
Micrometer or micron |
µ |
.000001 |
10 |
Nanometer |
nm |
.000000001 |
10 |
Angstrom |
Å |
.0000000001 |
10 |
From this table it is apparent that:
Measure |
Symbol |
Relative Volume |
Exponential Notation |
Liter |
L |
1 |
10 |
Deciliter |
dl |
.1 |
10 |
Millimeter |
ml |
.001 |
10 |
Microliter |
µ l |
.000001 |
10 |
Measure |
Symbol |
Relative Weight |
Exponential Notation |
Kilogram |
Kg |
1000 |
10 |
Gram |
g |
1 |
10 |
Milligram |
mg |
.0001 |
10 |
Microgram |
µ g |
.0000001 |
10 |
Molarity equals the number of moles of solute in 1 liter of solution. A mole is equal to the gram molecular weight (or formula weight) of the solute. Sodium Chloride (NaCl), for example has a formula weight of 58.43 (22.98 for Na and 35.43 for Cl). Thus, if 58.43 grams of NaCl are dissolved in 1 liter of water, the result would equal a 1 molar solution of NaCl. This is designated as 1 M NaCl, or as simple M NaCl.
We often deal with solutions of less volume than 1 liter, and the following should be noted:
1 M NaCl = 58.43 grams / liter = 58.43 mg/ml = 58.43 µ U g/µ U l.
A 0.002 M NaCl solution contains 0.002 moles of NaCl or 0.1168 grams (0.002 x 58.43) in one liter of solvent. Note that molar is abbreviated as M, but that there is no abbreviation for moles. The 0.002 M solution contains 0.002 moles (or 2 millimoles) or solute in one liter. A 0.002 M solution would contain 0.001 moles of solute in a half liter.
Note that chemical equations are always balanced via moles. Moreover, note that for dilutions of known concentrations, one can use the simple formula:
If you have a 0.002 M solution of NaCl and you wish to obtain 100 ml. of a 0.001 M solution,
Molarity is appropriate for use when chemical equations are to be balanced. When we deal with physical properties of solutions, molarity is not as valuable as a similar measurement of concentration, molality. For colligative properties of solutions (freezing point depression, boiling point elevation, osmotic pressure, density, viscosity), there is a better correlation between the property and molality.
Molality (designated with a lower case m) is equal to the number of moles of solute in 1000 gm of solvent. At first this may not appear any different from molarity, since a ml of water equals 1.0 gm. Indeed, for dilute solutions in water, there is little or no practical difference between a molar solution and a molal solution. In concentrated solutions, with temperature fluctuations and with changes in solvent, there is appreciable difference.
For example:
A 2 m (2 molal) solution of sucrose contains 684.4 gm of sucrose (twice the molecular weight or two moles of sucrose) dissolved in 1000 gm (approximately 1 liter) of water. The weight of this solution is 684.4 gm + 1000 gm or 1684.4 gm.
This solution (2 m sucrose) has a density of 1.18 gm/ml or 1180 gm/liter.
Since there are 1684.4 gms, division by the density (1180 gm/liter) would indicate that there are 1.43 liters of solution. That is, 684.4 gm of sucrose dissolved in 1000 gm of water would yield 1.43 liters of solution. This solution would contain 2 moles of sucrose, however and would have a molarity = 2 moles/1.43 liters or 1.40 M. So, a 2 m sucrose solution equals a 1.4 M sucrose solution.
There are three means of expressing concentration in the form of a percent figure:
For dilute solutions, these differences are not significant, but at higher concentrations, they are. Chemists (when they use Percent designations) usually use w/w. Biochemists and physiologists more often use w/v. Both use v/v if the solute is a liquid. It is important to distinguish among these alternatives.
Using ethanol as an example, consider a 20% solution of ethanol in water, mixed according to the three designations of w/w, w/v and v/v.
The three solutions are not the same. First, the density of alcohol is not equal to that of water, and thus conversion of g to ml is not equivalent. A 20% (w/w) solution of ethanol, for example, has a density of 0.97 g/ml and 20 gm of ethanol plus 80 gm of water would have a volume of 103 ml. The % (w/v) for this solution would be 20 gm ethanol / 103 ml, or 19.4% (w/v). Similarly, absolute ethanol has a density of 0.79 gm/ml and thus 20 ml of ethanol would weigh 15.8 gm. A 20% (v/v) solution would contain 15.8 gm of ethanol in 100 ml and be a 15.8% (w/v) solution.
So, for ethanol:
In cell biology, the most common use of Percent solution is as (w/v). In practice, these are simple solutions to mix. For a 20% (w/v) sucrose solution, for example, simply weigh 20 gm of sucrose and dissolve to 100 ml with water.
Unless specifically stated otherwise, solutions lacking the appropriate designation should be assumed to be (w/v).