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Molarity etc.

3 Ways to Measure Solution Strength.

Molarity = Moles of solute/Liters of solution.
Units: mol/L
Abbreviation: M

Example:

WHAT IS THE MOLARITY OF A SOLUTION THAT HAS 200 g
OF MgCl2 IN 1500 ml OF SOLUTION?
Remember in means 'divided by', so top and bottom of 1st factor-label cell must be used.

Factor-label set-up

200g MgCl2 1000mL 1 mol
1500mL 1L 95.21g MgCl2

= 1.4 M

* When mixing solutions. Weigh calculated mass of dry chemical. Add it to about 3/4 of the needed water. Then very carefully add water to exactly the needed amount. Usually in a volumetric flask.

Molality: moles of solute/ kilogram of solvent.
Units: mol/Kg
Abbreviation: m
(little m, not big M)

Example:

WHAT IS THE MOLALITY OF A SOLUTION THAT HAS 5.2
MOLES OF LEAD CHROMATE IN 2000 g OF WATER?

Factor label set-up

5.2 mol PbCrO4 1000g
2000g H2O 1 Kg

=2.6 mol/Kg or 2.6 m


Harder example:

WHAT IS THE MOLALITY OF A SOLUTION MADE BY
DISSOLVING 45.0 GRAMS OF DEXTROSE, C6H12O6, IN 500
GRAMS OF WATER?

45g Dextrose 1000g H2O  1 mol Dextrose
500g H20  1 Kg H20 180 g Dextrose

 = 0.5 m

Molality is really useful when solvent has a different density than 1.0 g/mL, like with organic solvents. Molality also controls freezing point depression, boiling point elevation.

Normality: equivalents of solute/liter of solution
Units:eq/liter
Abbreviation: N

Added step to factor label problems: # equivalents/mole, or number of positive charges/mole

Example:

HOW MANY GRAMS OF ZINC NITRATE WOULD
YOU NEED TO MAKE 1.7 LITERS OF 0.6 N SOLUTION?
Remember N really stands for 0.6 eq/L so it is a compound unit. Start the factor label problem with the single unit 1.7L !

Factor label set-up

1.7 L 0.6 equivalents 1 mol 189.4 g Zn(NO3)2
  1 L 2 equivalents 1 mol

                                                                       Zn is +2 so 2 positive charges or 2eq/mole of substance
 = 96.6 g
*This step in red is somewhat tricky.To find the equivalents per mole of substance look at the compound formula. Find the cation. (Zn2+) multiply by the subscript (unwritten 1), 2 x 1 = 2 equiv./mole of compound.

Why be normal?

Different compounds bring with them different numbers of positive charges. With phosphoric acid (H3PO4 you get 3 positive charges per mole! With HCl you only get 1. So if you had and an acid spill and were neutralizing it with NaOH (1 positive charge per mole) you would need 3 times as much to neutralize phosphoric acid as it would take to neutralize the hydrochloric acid. Normality is for neutralization.
2. Remember, identify the cation (+ ion, not part of the polyatomic anion though : ] ). Write the charge down. Multiply by the subscript. This gives you equivalents/mole of the compound.