Molarity
A measure of the concentration of a solution in moles per Liter solution. A solution
is a homogeneous mixture in which the particle size is so small that it cannot be
seen with a microscope and is uniformly distributed throughout the solution.
A solution is made of a solute dissolved in to a solvent. A solute is the substance that
is dissolved into the solvent. For example in a solution of salt water, the salt is the
solute and the water is the solvent. A solution may be a solid dissolved in a liquid, a
liquid dissolved in a liquid, a gas dissolved in a liquid, a gas dissolved in a gas.
Molarity is represented by a capital M.
M= mol/L
Molarity is moles divided by Liters
Example problems:
1. Find molarity of a solution that has 3g MgCl2 dissolved in 50mL water.
3g MgCl2/95g/mol MgCl2 = 0.0315mol MgCl2
0.0315 mol MgCl2/0.05L = 0.632 M MgCl2
2. A student has 190mL of 1.1M MgCl2 solution. Find the mass of MgCl2 dissolved
in solution.
M(L)= mol
(1.1)(0.19)=0.209 mol MgCl2
0.209 mol MgCl2 x 95g/mol MgCl2 = 19.86g MgCl2
3. How many liters of solution can be made of a 2M solution using 80g NaOH?
mol/M= L
80g NaOH/40g/mol NaOH = 2mol NaOH
2mol NaOH/2M = 1L solution
Dilution Problems:
To dilute a solution is to make it less concentrated by adding more solvent. Often
times in chemistry we need to make a dilute solution from a highly concentrated
original solution. The equation is as follows:
M1V1=M2V2
Where M1 and V1 are the molarity and volume of the original solution and M2 and V2
are the molarity and volume of the dilute solution.
Example:
1. How would you prepare 2L of a 0.25M NaOH solution from 1M stock
solution?
*the stock solution is the more concentrated solution you have in stock or
that you are starting with. You are figuring out how much of the concentrated
solution to pour into a beaker and then how much water you will add to
dilute it to the concentration and volume you want.
M1V1=M2V2
(1M)(V1) = (0.25M)(2L)
V1= 0.5L
Start with 0.5L of 1MNaOH and add 1.5L of water
(The amount of water added is the difference between the amount of original
solution you start with and the total volume of solution. In this case take the 2L of
final solution and subtract the 0.5L you will start with in the beaker. Therefore 1.5L
of water will be added to the 0.5L of 1M NaOH in the beaker.)