Math 1010
Molarity Project
Math Background:
The mathematics used in concentration calculations stem from the relationship between
amount (percentage), rate (percent) and base. Recall, A  r  B , amount equals rate
times base. So whether we are computing the tip on the cost of a meal, the discount on
an item on sale, or the amount of pure substance in a solution, you would multiply the
rate (percent) times Base. This basic formula can be “solved” for any of the specific
A
A
quantities, r  or B  . When doing calculations you must be aware of how units
B
r
are behaving. The desired unit must be the outcome of the computation.
Chemistry Background:
When referring to the concentration of a solution, we generally mean the amount of
solute contained in a certain amount of solution. By varying the amount of solute
dissolved in a solution the concentration changes. The strength of the concentration is
given in terms of a numerical value. The percentages will have units of measurement
using combinations of mass, volume or moles. For example, volume/volume percent,
weight (mass)/weight percent, weight/volume percent, or molarity.
Volume percent is used when two liquids are mixed.
Volume percent 
Example:
volume of solute
100%
volume of solution
If a solution has a (v/v%) concentration of 32.5%. How many mL of solute
are present in 350 mL solution?
Here you would use the formula amount of pure substance (volume of solute) equals
concentration times volume (solution). The concentration of 32.5% means 32.5 mL
solute contained in 100 mL solution.
A  C V  A 
32.5 mL solute
 350 mL solution  A  113.75 mL solute
100 mL solution
Weight/volume percent is used when a mass in grams is dissolved in a liquid measured
in milliliters.
weight / volume percent 
mass of solute ( g )
100%
volume of solution (mL)
Example:
If a solution is 37.5% (m/v) glucose, how many mL of solution must be
administered to the patient if the doctor orders 9 grams of glucose?
The concentration tells us there is 37.5 grams glucose in 100 mL solution. This is an
equivalence relationship, a conversion factor that can be used to set up the
computation. The desired outcome is mL of solution and the doctor ordered 9 grams of
glucose. The units of mass must cancel out, so the “conversion factor”
100 mL solution
37.5 g glucose
100 mL solution can be set up as 37.5 g glucose depending on what units are needed
in the desired outcome.
Therefore,
100 mL solution
 9 g glucose  24 mL solution
37.5 g glucose
You would administer 24 mL of the solution to the patient.
In the formula A  C V , the goal is to solve for V, so divide by the concentration, hence
A
V
will give the volume to administer.
C
V
9 g glucose
37.5 g glucose
100 mL solution . To get the outcome, recall how fractions are divided.
We often express concentrations in terms of Molarity, which is the number of moles of
solute dissolved in a liter of solution.
Molarity 
moles of solute
volume of solution ( L)
The units would be moles of solute per liter of solution.
Example:
A solution that contains 0.7 moles of solute per 5 L solution would have molarity:
Molarity 
0.7 moles solute
moles
 Molarity  0.14
 Molarity  0.14 M
5 L solution
L
Dilutions
In a lab, often you are working with stock solutions and need to dilute the concentration
in order to perform an experiment. The important concept to remember is that diluting a
substance does not change the amount of that substance. So the number of moles of
the substance before the dilution is the same as the number of moles after the dilution
(conservation of matter). Thinking about what that means, If you are making orange
juice and have one can of orange juice and add three cans of water, you have a solution
1 can OJ
that is
 25% OJ now if you dilute it by adding another can of water you now
4 total cans
have
1 can OJ
 20% OJ . The concentration decreased, but you still have only one
5 total cans
can of orange juice in the solution. The total amount of pure substance did not change,
a formula to reflect this concept is CiVi  C f V f . In words initial concentration times initial
volume equals final concentration times final volume.
Example: You are a intern in a lab and your supervisor has just asked you to prepare
500mL of 3M HCl for the experiment. The stockroom cabinet contains "Stock
Solutions". In it there is a 2.5L bottle labeled "11.6M HCl". The concentration of the HCl
is 11.6M. Please prepare a flask containing 500 ml of a 3 M solution, relabeling it with
its precise molarity. How do you prepare this solution?
The formula to use is CiVi  C f V f because the total number of moles will be the same.
How much stock do you use? Solve for Vi .
Initial: Ci  11.6
moles HCl
, Vi  ? L HCl
L
Final: C f  3
moles HCl
, V f  0.5 L HCl
L
Note 500 mL = 0.5 L
Therefore CiVi  C f V f becomes 11.6
moles HCl
moles HCl
 Vi L HCl  3
 0.5 L
L HCl
L
(conservation of mass)
This simplifies as 11.6 Vi moles HCl  1.5 moles HCl solving for final volume (divide both
sides by 11.6) notice moles cancel, this gives V f  0.13 L HCl . Therefore, take 130 mL
of the 11.6 M stock solution and add enough water so that the total volume is 500 mL.
Lab Activity:
You are working as an intern in a biotechnology lab. Teddi, your supervisor, has just
asked you to perform the following tasks:
Please describe your complete procedure and the key quantities you
measure. Points are based on whether or not you explain your procedure
in sufficient detail for us to know what you did.
1. Prepare 500 mL of a 5% (w/v) solution of NaCl.
We used cross multiplication for this problem. The answer will be 25mL NaCl (my
work is on the written copy of my paper)
2. Prepare 30 g of a 25% (w/w) solution of NaCl.
We used cross multiplication for this problem. The final answer is 7.5 g of NaCl.
(Work is on scanned copy)
3. Prepare 200 mL of a 0.8 M solution of CaCl2.
We used cross multiplication and then we converted the answer to Moles. The final
answer is 0.16 Moles. (Work is on scanned copy)
4. Prepare 450 mL of a 35% stock solution using a 70% stock solution and a 20%
stock solution.
We used a table to set up the problem, and then we used elimination to find the
solution. The final solution is 135mL of 70% and 315 mL of 20 %.( Work is on
scanned copy)
5. Prepare 500 mL of 2.00 M buffer solution for an experiment. In the Stock
Solutions cabinet you will find a 2.50 liter bottle containing 5.6 M buffer
solution.
Once we divided 1000 by 5.6 we had to convert our answer into Miles. To do this we
divided our answer by 1000 and our final answer was 0.17857 or when rounded 0.18
Miles. (Work is on scanned copy)
6. What volume of a 40% acidic solution contains 4.2 grams of acid?
For number six we first had to convert grams into mL. Afterwards we got 42/4; our
final answer was 10.5L. (Work is on scanned copy)
Reflection:
1. What are you trying to do in this lab?
We are using mathematical for moles to find the answer to the problem.
2. What technique or basic principle are you using or testing?
We are using the percent formula
3. Describe any source of experimental error. How can they be avoided?
There many errors that can be made for example humane error in calculation, error in
conversions error in use of formula ect….
4. What problems did you encounter in completing this assignment?
We had to know what formulas to use. Also we had to translate the word problem
properly. We had to remember to convert measurements when needed.
5. How is this assignment related to other coursework?
We had to remember formulas and techniques also we practiced in our homework.