l.OrmA +1000 // OsmM _?20 Your Turn In lab, you will use a stock solution that i
ID: 145149 • Letter: L
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l.OrmA +1000 // OsmM _?20 Your Turn In lab, you will use a stock solution that is 2 mM of KMnOs in a serial dilution to make 6 mL of each of the include the units in your answers, where appropriate followving concentrations 1.0 mM 05m250TM25M 625 M. 313 JM. 156 M. Remember to 1. What is the dilution factor for this serial dilution?a 26012-2 ete M 250 UN 25 uM. 62.5 uM. 31.3 uM, 15.6 UM Remember to 1000 15v0 2 2. What is V2 for this serial dilution? 3. Calculate Vi for this serial dilution. Show your calculations. 2 4. What volume of water will you add to each tybe? Transfer vour answers for Questions 1-3 to Part IV of the Lab ProceduresExplanation / Answer
1) Initially, 1 mM solution is prepared from 2 mM solution of KMnO4.
Initial Concentration C1= 2 mM KMnO4
Final concentration = 1 mM KMnO4
Initial volume= V1=?
Final volume= V2= 6 mL
C1V1= C2V2
2 mM KMnO4 X V1= 1 mM KMnO4 X 6 mL
V1= 1 X 6/2= 6/2= 3 mL
Since, 3 ml of 2 mM solution is added to 3 mL of distilled water to obtain a 6 ml solution of 1 mM KMnO4 solution, the first dilution is a 1:2 dilution.
Hence, dilution factor of first dilution =2.
2) 0.5 mM KMnO4
C1= 1 mM KMnO4
V1=?
C2= 0.5 mM KMnO4
V2= 6 mL
C1V1=C2V2
V1= C2V2/C1= 0.5 mM KMnO4 X 6 mL/1 mM KMnO4 = 3/1= 3 mL
Again, this is 1: 2 dilution of 1 mM KMnO4 solution. Dilution factor is 2.
3) 250 micromolar KMnO4
1 mM KMnO4= 1000 micromolar KMnO4
0.5 mM KMnO4= 500 micromolar KMnO4
C1= 500 micromolar KMnO4
V1=?
C2= 250 micromolar KMnO4
V2= 6 mL
C1V1=C2V2
V1= C2V2/C1= 250 micromolar KMnO4 X 6 mL/ 500 micromolar KMnO4 = 250 *6/500= 3 mL
Again, this is 1: 2 dilution of 500 micromolar KMnO4 solution. Dilution factor is 2.
4) 125 micromolar KMnO4
C1= 250 micromolar KMnO4
V1=?
C2=125 micromolar KMnO4
V2= 6 mL
C1V1=C2V2
V1= C2V2/C1= 125 micromolar KMnO4 X 6 mL/ 250 micromolar KMnO4 = 125*6/250= 3 mL
Again, this is 1: 2 dilution of 250 micromolar KMnO4 solution. Dilution factor is 2.
5) 62.5 micromolar KMnO4
C1= 125 micromolar KMnO4
V1=?
C2=62.5 micromolar KMnO4
V2= 6 mL
C1V1=C2V2
V1= C2V2/C1= 62.5 micromolar KMnO4 X 6 mL/ 125 micromolar KMnO4 = 62.5*6/125= 3 mL
Again, this is 1: 2 dilution of 125 micromolar KMnO4 solution. Dilution factor is 2.
6) 31.3 micromolar KMnO4
C1= 62.5 micromolar KMnO4
V1=?
C2=31.3 micromolar KMnO4
V2= 6 mL
C1V1=C2V2
V1= C2V2/C1= 31.3 micromolar KMnO4 X 6 ml/ 62.5 micromolar KMnO4=31.3*6/62.5= 3 ml
Again, this is 1: 2 dilution of 62.5 micromolar KMnO4 solution. Dilution factor is 2.
7) 15.6 micromolar KMnO4
C1= 31.3 micromolar KMnO4
V1=?
C2=15.6 micromolar KMnO4
V2= 6 mL
C1V1=C2V2
V1= C2V2/C1= 15.6 micromolar KMnO4 X 6 mL/ 31.3 micromolar KMnO4= 15.6*6/13.3= 3 mL
Again, this is 1: 2 dilution of 31.3 micromolar KMnO4 solution. Dilution factor is 2.
Below are the answers that are required in lab report:
1) Dilution Factor of this serial dilution is= 2 (as all are 1: 2 dilutions)
Dilution factor= Final volume/Initial volume= 6mL/3mL= 2
2) V2= 6 mL (as final volume required to be prepared is mL)
3) V1= 3 mL
Example given: 1 mM KMnO4 prepared from 2 mM KMnO4
C1V1= C2V2
2 mM KMnO4 X V1= 1 mM KMnO4 X mL
V1= 1 mM KMnO4/6 mL/ 2 mM KMnO4= 1 *6/2= 3 mL
Dilutions prepared are 1:2 (dilution factor is 2). Hence, it will always be 3 mL of the previous concentration with 3 mL of distilled water to give 6 mL final volume of required concentration. (Dilution 1:2= 1 mL of stock + 1 mL of diluent).
4) There is equal volume of distilled water added in a 1: 2 dilution. Hence, volume of distilled water added = 3 mL
First and second answers are right. The third answer formula is C1V1=C2V2. This affected the fourth answer as well.
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