A complete description of a solution states what the solute is and how much solute is dissolved in a given amount of solvent or solution. The quantitative relationship between solute and solvent is the concentration
of the solution. This concentration may be expressed using several different methods, as discussed next.
A. Concentration by Mass
The concentration of a solution may be given as the mass of solute in a given amount of solution, as in the following statements: The northern part of the Pacific Ocean contains 35.9 g salt in each 1000 g seawater. The North Atlantic Ocean has a higher salt concentration, 37.9 g salt/1000 g seawater.
B. Concentration by Percent
The concentration of a solution is often expressed as
percent concentration by mass or percent by volume of solute in solution. Percent by mass is calculated from the mass of solute in a given mass of solution. A 5%bymass aqueous solution of sodium chloride contains 5 g sodium chloride and 95 g water in each 100 g solution.
Percent by mass  =  mass of solute mass of solution 
X  100% 
Example: How many grams of glucose and of water are in 500 g of a 5.3% bymass glucose solutionSolution We know that 5.3% of the solution is glucose: The remainder of the 500 g is water: 
If both solute and solvent are liquids, the concentration may be expressed as percent by volume. Both ethyl alcohol and water are liquids; the concentration of alcoholwater solutions is often given as percent by volume. For example, a 95% solution of ethyl alcohol contains 95 mL ethyl alcohol in each 100 mL solution.
Percent by volume  =  volume of solute volume of solution 
X  100% 
Example: Rubbing alcohol is an aqueous solution containing 70% isopropyl alcohol by volume. How would you prepare 250 mL rubbing alcohol from pure isopropyl alcohol?Solution We know that 70% of the volume is isopropyl alcohol:
To prepare the solution, enough water is added to 175 mL isopropyl alcohol to form 250 mL solution.:

Because the density of liquids changes slightly as the temperature changes, a concentration given in percent by mass is accurate over a wider range of temperatures than is a concentration given in percent by volume. Sometimes a combination of mass and volume is used to express the concentrationthe mass of solute dissolved in each 100 mL solution. Using this method, a 5% (wt/vol) solution of sodium chloride contains 5 g sodium chloride in each 100 mL solution.
C. Concentration in Parts per Million (ppm)
and Parts per Billion (ppb) The
terms (ppm) and parts per billion (ppb) are encountered more and more frequently
as we become aware of the effects of substances present in trace amounts in
water and air, and as we develop instruments sensitive enough to detect substances
present in such low concentrations. In discussing mass, parts per million means
concentration in grams per 10^{6} grams, or micrograms per gram. In
discussing volume, parts per million may mean milliliters per cubic meter, or
the mixed designation of milligrams per cubic meter. For parts per billion,
the general trend is toward the use of micrograms per liter when discussing
water contaminants, micrograms per cubic meter for air, and micrograms per kilogram
for soil concentrations.
D. Concentration in Terms of Moles
The concentration of a solution may be stated as molarity
(M), which is the number of moles of solute per liter of solution or the number of millimoles
(mmol) (1 millimole = 10^{3} mole) per milliliter of solution.
Molarity (M)  =  moles solute volume (liter) solution 
=  millimoles solute milliliter solution 
A 6 M (say "six molar") solution of hydrochloric acid contains 6 mol hydrochloric acid in 1 L solution.
The molarity of a solution gives a ratio between moles of solute and volume of solution. It can be used as a conversion factor between these two units in calculations involving solutions. As a conversion factor, it can be used two ways:
Example: How many moles of hydrochloric acid are in 200 mL of 0.15 M HCl?Solution Wanted: ? mol HCl Given 200 mL of 0.15 M HCl Conversion Factors Equation Answer 0.30 mol HCl Note that, each time the volume of a solution is stated, the concentration of the solution is given. This form may look confusing, but without this marking it is easy to forget which solution you are referring to. 
Example: What mass of sodium hydroxide (NaOH) is needed to prepare 100 ML of 0.125 M sodium hydroxide?Solution Wanted: ? g NaOH Given 100 mL of 0.125 M NaOH Conversion Factors 1 L of 0.125 M NaOH contains 0.125 mol NaOH that is, Equation Answer 0.500 g NaOH 
Example: What volume of 3.25 M sulfuric acid is needed to prepare 0.500 L of 0.130 M H_{2}SO_{4}?Solution We are to prepare 0.500 L of 0.130 M H2SO4 by adding an amount of water to an amount of 3.25 M H2SO4. The moles of sulfuric acid in the final (more dilute) solution will be the same as the moles of sulfuric acid in the portion of the more concentrated solution. We can calculate the moles of sulfuric acid in the final dilute solution: This answer gives the moles of acid needed. We can calculate the volume of 3.25 M H_{2}SO_{4} that would contain 0.065 mol H_{2}SO_{4}. This answer gives the volume of concentrated acid that conatins the moles of acid needed for the dilute solution. This volume of 3.25 M H_{2}SO_{4} would be dissolved in 480 mL (500 mL  20 mL) water to prepare 0.500 L of 0.130 M H_{2}SO_{4}. This problem is diagramed in the figure.

Example: What volume of 6.39 M sodium chloride contains 51.2 mmol sodium chloride?Solution Wanted Given Conversion factors Equation

Example: How do we prepare 75.0 mL of 0.96 M sulfuric acid from 18 M acid?Solution We are to prepare 75.0 mL of 0.96 M sulfuric aicd by diluting 18 M sulfuric acid with water. We can calculate the millimoles of sulfuric acid in the final solution:
We can calculate the volume of 18 M ? mmol H_{2}SO_{4} that will contain 72 mmol H_{2}SO_{4}:
The solution would be prepared by adding 4.0 mL of 18 M H_{2}SO_{4} to about 50 mL of water and then dilutint that solution to exactly 75.0 mL. 
Table 11.3 lists several of the commonly used ways of expressing concentrations.
Solute  Solvent  Solution  Comments  

Percent by weight  ? g  +  ? g  100 g 

accurate, independent of temperature  
Percent by volume  ? mL  +  ? mL  100 mL 

used when solute is liquid; concentration varies slightly with tenperature  
Percent, weight/volume  ? g  +  100 mL 

used when solute is liquid; concentration varies slightly with tenperature  
Molarity (M)  moles 

1 liter  }  used in chemical calculations  
Millimole/liter  10^{3} mole 

1 liter  
Millimole/milliliter  10^{3} mole 


10^{3} liter  
Parts per million (ppm)  mg 

kg  }  used in environmental studies  
Parts per billion (ppb)  µg 

kg  
