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MiniUnit ~ Solutions (Chapter 15)
Introduction and Definitions (Page 451)
Solutions are important to chemists because much of the chemistry in the lab, the environment,
and within living organisms, occurs in solutions.
A brief review of terms:
Mixture =
two or more substances physically combined. Homogenous mixtures have
similar properties throughout, whereas heterogenous mixtures’ properties vary.
Solution =
a homogenous mixture. Examples below:
Solution
Salt Water
Carbonated Water
Gold Jewelry
Air
Solute
Solvent
Solute is present in ______________ amount. Solvent is present in ______________ amount
Solution = Solute + Solvent in other words:
The solute is dissolved in the solvent!
Concentration and Molarity (Sections 15.2, 15.4)

For solutions, chemists use the idea of ______________________ to describe the amount of
solute in a given amount of solution.

Concentrations are always RATIOS:

“Molarity” is a common method of expressing
the concentration of a solute dissolved in a solution.
Molarity, abbreviated by M, is defined as:
amount of solute
amount of solvent or solution
Note: the volume MUST be in Liters!

Sample Molarity Calculation:
What is the molarity (M) of a 250. mL solution
containing 9.46 g of CsBr?
9.46 g CsBr x 212.81 g/mole =
M = moles solute =
L solution

Sample Molarity Problem:
Calculate the number of grams of NaCl needed to
make 125 mL of a 3.78 M solution.
Picture:
The above problem is an example of how to make a molar solution from a solid. Now, we turn
to preparation of a molar solution by dilution.
Dilutions (Section 15.5)
One of the most common lab techniques is the dilution of a more concentrated solution to make a
less concentrated solution. The basic idea is for the procedure is to remove a “small” portion of
the concentrated solution (often referred to as the “stock”), place it into a volumetric flask, and
then fill the flask to a mark with water. The addition of the water “dilutes” the concentrated
stock to the new molarity.
To calculate the amount of stock to remove and the volume of your new solution, use the
equation:
MCVC = MDVD
where M = molarity, V = volume, C = concentrated, and D = dilute
Example:
Calculate the volume of solution needed to prepare 500 mL of a 3.0 M HCl
solution from the 12.0 M stock.
MC = larger molarity = 12.0
MD = smaller molarity = 3.0
VC = ?
VD = 500. mL (doesn’t have to be in L)
VC = MD x VD = 3.0 x 500 = 125 mL
MC
12.0
Equipment:
Pipet
Volumetric
Picture:
Pipet 125 mL from stock
into 500 mL Volumetric.
Fill to the 500 mL mark
with water.
Note how the dilution factor of 4 (from 12M to 3M) is the same as the factor of water
addition: 125 mL stock x 4 will give 500 mL solution total!
Practice:
Calculate and draw a picture for the preparation of 100. mL of 2.0 M HNO3
solution from a 16 M HNO3 stock.
Picture: