Microbiology lab 2014
Assignment 1
Due date: Sept. 17
Part I: Problems. Solve the following problems. You are not required to show your calculations.
Only submit your final answers. (1 point/question)
1. What is the molarity of a solution of ammonium chloride prepared by diluting 155.0 mL of a
2.79 M NH4Cl solution to 2.5 L?
Initial Concentration = 2.79 M of NH4Cl
Initial Volume = 0.155L
Final Volume = 2.5L
Final Concentration = ? of NH4Cl
C1= 2.79M; V1= 0.155L; C2= ?; V2= 2.5L;
C2 = C1V1/V2
C2 = 0.155L * 2.79M / 2.5L
C2 = 0.173M
2. A student takes a sample of a KOH solution and dilutes it by adding 100.00 mL of water.
The student determines that the diluted solution is 0.016 M KOH. Given that the
concentration of the original solution was 2.09 M. What was the volume of the original
sample?
Initial Concentration = 2.09M of KOH
Initial Volume = X mL
Final Volume = X + 100mL
Final Concentration = 0.016M of KOH
C1= 2.09M; V1= X mL; C2=0.016M; V2= (X + 100) mL;
C1V1 = C2V2
2.09M * X ml = 0.016M * (X + 100) mL
2.09X = 0.016X + 1.6
2.09X – 0.016X = 1.6
2.074X = 1.6
X = 1.6/2.074
X=0.77mL
The initial concentration is 0.77mL.
3. A microbiologist wants to prepare a stock solution of H2SO4 so that samples of 25.0 mL will
produce a solution with a concentration of 0.50 M when added to 100.0 mL of water. What
should the molarity of the stock solution be?
Initial Concentration = ? M of H2SO4
Initial Volume = 25mL
Final Volume = 25mL (initial volume) + 100mL (added water) = 125mL
Final Concentration = 0.5M of H2SO4
Microbiology lab 2014
C1= ? M; V1= 25mL; C2=0.5M; V2= 125mL;
C1 = C2V2 / V1
C1 = 0.5M * 125mL / 25mL
C1 = 2.5M
4. What volume of water should be added to 1.19 mL of an 8.0 M acetic acid solution in order
to obtain a final concentration that is 1.5 M acetic acid?
Initial Concentration = 8.0M of acetic acid
Initial Volume = 1.19mL
Final Volume = ? mL
Final Concentration = 1.5M of acetic acid
C1= 8.0M; V1= 1.19mL; C2=1.5M; V2= ? mL;
V2 = C1V1 / C2
V2 = 1.19mL * 8M / 1.5M
V2 = 6.35mL
Volume of added water = 6.35mL – 1.19mL = 5.14mL H2O
5. Three solutions "A", "B", and "C" are mixed to obtain the following ratio: A:B:C = 1:2:27.
What are the dilution factors for each of these compounds?
A = 1/30, therefore A (dilution factor) = 30/1 = 30x
B = 2/30, therefore B (dilution factor) = 30/2 = 15x
C = 27/30, therefore C (dilution factor) = 30/27 = 1.11x
6. A microbiologist has three microbial cultures: E.coli at a density of 1 X 108 cells/mL, B.
subtilis at a density of 5 X 109 cells/mL, and P. notatum at a density 2.5 X 109 cells/mL.
From these, he wishes to prepare a single mixture containing 5 X 106 cells/mL of E.coli,
1.25 X 108 cells/mL of B. subtilis, and 1 X 105 cells/mL of P. notatum in a final volume of
10mL of media. What volume of media and of each of the original cultures should be used
to achieve this?
E.Coli:
Initial Concentration = 1 x 108 cells/uL
Initial Volume = ? mL
Final Volume = 10mL
Final Concentration = 5 x 106 cells/uL
C1= 1 x 108 cells/uL; V1= ?mL; C2= 5 x 106 cells/uL; V2= 10mL;
V1 = C2V2 / C2
V1 = 1 x 108 cells/uL * 10mL / 5 x 106 cells/uL
V1 = 0.5mL
Volume of initial E.Coli culture = 0.5mL
Microbiology lab 2014
B.Subtilus:
Initial Concentration = 5 x 109 cells/uL
Initial Volume = ? mL
Final Volume = 10mL
Final Concentration = 1.25 x 108 cells/uL
C1= 5 x 109 cells/uL; V1= ?mL; C2= 1.25 x 108 cells/uL; V2= 10mL;
V1 = C2V2 / C2
V1 = 5 x 109 cells/uL * 10mL / 1.25 x 108 cells/uL
V1 = 0.25mL
Volume of initial B.Subtilus culture = 0.25mL
P.notatum:
Initial Concentration = 2.5 x 109 cells/uL
Initial Volume = ? mL
Final Volume = 10mL
Final Concentration = 1 x 105 cells/uL
C1= 2.5 x 109 cells/uL; V1= ?mL; C2= 1 x 105 cells/uL; V2= 10mL;
V1 = C2V2 / C2
V1 = 2.5 x 109 cells/uL * 10mL / 1 x 105 cells/uL
V1 = 0.0004mL
Volume of initial P.notatum culture = 0.0004mL
Volume of culture media = 10mL – (0.5mL + 0.25mL + 0.0004mL) = 9.25mL
7. 7 parts of water are added to 1 part of a 3.8 M solution of FeSO4. What is the molarity of the
diluted solution?
7 Parts Water
1 Part FeS04
C1= 3.8M; V1= 1 part; C2= ?M; V2= 8parts;
C2 = C1V1 / V2
C2 = 3.8M * 1 part / 8 parts
C2 = 0.475M
8. A microbiologist prepares 480 mL of a 2.50 M solution of K2Cr2O7 in water. A week later, 39
mL of water has evaporated. How many moles of K2Cr2O7 are left and what is the new
molarity of the solution?
Initial Concentration = 2.5M
Initial Volume = 480 mL
Microbiology lab 2014
Final Volume = Initial Volume – Evaporated water = 480mL – 39mL = 441mL
Final Concentration = ?M
C1= 2.5M; V1= 480mL; C2= ?M; V2= 441mL;
C2 = C1V1 / V2
C2 = 2.5M * 480mL / 441mL
C2 = 2.72M
Initial # of K2Cr2O7 moles = Final # of K2Cr2O7 moles
2.5M * 0.48L = 2.72M * 0.441L
1.2 moles = 1.2 moles
Final # of K2Cr2O7 moles = 1.2 moles
9.
A chemical test has determined the concentration of 100 mL of a solution of an unknown
substance to be 2.41 M. The solution is totally evaporated, leaving 9.56 g of crystals of the
unknown solute. What is the molar mass of the unknown substance?
Volume = 0.1L
Concentration = 2.41M
Therefore, # of moles = 0.1L * 2.41M = 0.241 moles
Molar mass = Mass of crystals / # of moles
Molar mass = 9.56g / 0.241moles
Molar mass = 39.67 g/moles
10. How many milliliters of a 250mg/mL chloramphenicol solution are needed for a 4 g dose?
Volume of chloramphenicol (250mg/mL) for 4g dose = Masse / Concentration
Volume of chloramphenicol (250mg/mL) for 4g dose = 4000mg / 250mg/mL
Volume of chloramphenicol (250mg/mL) for 4g dose = 16mL
11. A pharmacist hands you a 1.5L bottle of a 20% (m/v) NaCl solution and asks that you mix it
with sterile water to make as much of a 0.1M solution as possible. How much sterile water
would you use? (MW of NaCl 58g/mole)
Convert NaCl concentration from % (m/v) to molarity:
20% NaCl = 20g/100mL
20% NaCl = 200g/1000mL
# of moles / 1L = Mass in 1000mL / MM
# of moles / 1L = 200g / 58g/mole
# of moles / 1L = 3.45 moles
Therefore, molarity of 20% NaCl = 3.45M
Initial Concentration = 3.45M
Initial Volume = 1.5L
Microbiology lab 2014
Final Volume = ?L
Final Concentration = 0.1M
C1= 3.45M; V1= 1.5L; C2= 0.1M; V2= ?L;
V2 = C1V1 / C2
V2 = 3.45M * 1.5L / 0.1M
V2 = 51.75L
Needed sterile water = Final Volume – Initial Volume
Needed sterile water = 51.75L – 1.5L
Needed sterile water = 50.25L
12. The following dilutions were performed to determine the concentration of bacteria in a
culture. What was the concentration of bacteria in the stock?
2 mL
1 mL
5 mL 10 mL
0.1 mL
Stock 8 mL
4 mL
15 mL
150
colonies
10 mL
Dilution 1 = 2mL / 10mL = 1/5
Dilution 2 = 1mL / 5mL = 1/5
Dilution 3 = 5mL / 20mL = 1/4
Dilution 4 = 10mL / 20mL = 1/2
Serial dilution = 1/5 * 1/5 * 1/4 * 1/2 = 1/200
Therefore, the dilution factor is = 200/1 = 200x
Concentration = Dilution factor * # of Colonies on plate * factor to bring amount
plated to 1mL
Concentration = 200 * 150 * 10
Concentration = 3 * 105 cells/mL
13. 52mL of sodium cyanide poison leaked from a bottle, which originally contained 100mL, into
a bucket of water containing 428mL of water. The concentration of poison in the bucket
was found to be 0.85 M after the leak. If the molecular weight of sodium cyanide is
49g/mole, how many grams of sodium cyanide remain in the bottle?
Volume of new cyanide solution = 52mL of original solution + 428mL of water
Volume of new cyanide solution = 480mL
Initial Concentration = ?M
Initial Volume = 52mL
Final Volume = 480mL
Final Concentration = 0.85M
Microbiology lab 2014
C1= ?M; V1= 52mL; C2= 0.85M; V2= 480mL;
C1 = C2V2 / V1
C1 = 0.85M * 480mL / 52mL
C1 = 7.85M
Therefore, concentration of initial cyanide solution = 7.85M
# of moles left in the initial bottle = volume left * molarity
# of moles left in the initial bottle = 0.048L * 7.85M
# of moles left in the initial bottle = 0.3768 moles
Mass of cyanide left in the initial bottle = # of moles * MM
Mass of cyanide left in the initial bottle = 0.3768 moles * 49g/mole
Mass of cyanide left in the initial bottle = 18.4632g
14. Concentrated hydrochloric acid has a concentration of 37.7% (m/m). What is its molar
concentration? (The density of the solution is 1.19 g/mL and MW of HCL: 36g/mole)
Concentration 37.7% (m/m) = 37.7g HCl / 100g solution
100g solution in mL = mass / density
100g solution in mL = 100g / 1.19g/mL
100g solution in mL = 84mL
Concentration 37.7% (m/m) = 37.7g HCl / 84mL solution
Mass of HCl in 1L of solution :
1000mL / 84mL = 11.9
(37.7g HCl / 84mL solution) * 11.9 = 448.63g / 1000mL
Molarity = Masse of HCl in 1L / MM
Molarity = 448.63g / 36 g/mole
Molarity = 12.5M
15. You wish to prepare a tetracycline solution of 1mg/ml in ethanol from a 10mg/ml stock
solution of tetracycline. Given that you have 10ml of the tetracycline stock solution and 5 ml
of ethanol, what is the maximum volume of the diluted solution that you can prepare?
First find which reagent is limiting, in this case it’s the ethanol.
Initial Concentration = 10mg/mL
Initial Volume = XmL of 10mg/mL
Final Volume = XmL of 10mg/mL + 5mL of ethanol
Final Concentration = 1mg/mL
C1= 10mg/mL; V1= XmL; C2= 1mg/mL; V2= X + 5 mL;
C1V1 = C2V2
Microbiology lab 2014
10mg/mL * X mL = 1mg/mL * (X + 5) mL
10X = 1X + 5
9X = 5
X = 5/9
X = 0.556mL
Therefore,
Volume of 1mg/mL of tetracycline produced = X + 5 mL
Volume of 1mg/mL of tetracycline produced = 0.556ml + 5mL
Volume of 1mg/mL of tetracycline produced = 5.556mL