THE ACTUAL WEIGHING OF REACTANTS OFTEN PROVES IMPRACTICAL OR INCONVENIENT. THIS IS ESPECIALLY TRUE IF REACTANTS ARE GASES, LIQUIDS, OR VERY REACTIVE.
             HOW CAN WE DELIVER KNOWN WEIGHTS OF REACTANTS WITHOUT WEIGHING THEM FIRST?
             CONSIDER: HNO3 + NaOH = NaNO3 + H2O
             NEITHER NITRIC ACID, NOR SODIUM HYDROXIDE CAN BE WEIGHED EASILY. WHY?
             HOWEVER, SOLUTIONS OF KNOWN AMOUNT OF SOLUTE PER UNIT VOLUME CAN BE DELIVERED IN KNOWN VOLUME TO GIVE PRECISELY-KNOWN AMOUNTS OF REACTANTS. WHY?
             BECAUSE: (MOLES/LITER) X LITERS = MOLES
             IF YOU KNOW THE NUMBER OF MOLES CONTAINED IN ONE LITER, YOU CAN CALCULATE THE MOLES CONTAINED IN ANY MEASURED VOLUME OF THAT SOLUTION.
             TYPICALLY, A FLASK CALLED A VOLUMETRIC FLASK IS USED TO PREPARE A SOLUTION OF KNOWN CONCENTRATION,
             (MOLES SOLUTE/LITERS SOLUTION). SOLUTE IS WEIGHED AND ADDED TO THE FLASK. THEN THE SOLVENT IS ADDED TO MAKE UP A KNOWN FIXED VOLUME OF SOLUTION.
             THE AMOUNT OF SOLUTE PER UNIT VOLUME SOLUTION(TYPICALLY MOLES/LITER) IS CALLED THE CONCENTRATION OF THE SOLUTION. IT HAS TO BE DEFINED QUANTITATIVELY. THIS CAN BE DONE IN SEVERAL WAYS. WE WILL USE THE DEFINITION MOST COMMONLY USED IN ANALYSIS, MOLARITY.
             THE MOLARITY IS GIVEN THE SYMBOL, M.
             IF A SOLUTION IS 1.000 M, 1.000 LITERS OF SOLUTION CONTAIN EXACTLY 1.000 MOLES OF SOLUTE.
             THE USEFULNESS OF THE CONCEPT REVOLVES AROUND THE FACT THAT
             YOU MUST KNOW HOW TO PREPARE DILUTE SOLUTIONS FROM CONCENTRATED ONES.
             IN MOST LABORATORIES REAGENTS ARE PURCHASED IN CONCENTRATED FORM DUE TO SHIPPING AND PACKAGING COSTS . THEY ARE RARELY USED AS DELIVERED. RATHER THEY ARE DILUTED FOR USE AS NEEDED. IF YOU EVER WORK IN A LABORATORY YOU MUST KNOW HOW TO PREPARE DILUTE SOLUTIONS FROM MORE CONCENTRATED ONES.
             ACS-CERTIFIED REAGENT-GRADE HCl IS SOLD IN BOTTLES CONTAINING 12.1 M SOLUTION.
             8.26 mL OF CONCENTRATED HCl WHEN ADDED TO
             MAKE ONE LITER OF SOLUTION RESULTS IN A SOLUTION WHICH IS 0.100 M.
             IT IS A LITTLE EASIER TO MAKE UP ...