Starting with a stock solution of 4 mM KMnO4, you plan to make 16 mL of...
Starting with a stock solution of 4 mM KMnO4, you plan to make 16 mL of each of the following concentrations of KMnO4: 1.0 mM, 250 μM, 62.5 μM , and 15.6 μM. What is the V1, expressed in mL, for this dilution series? Report your answer to 2 decimal places.
Homework Answers
here M1= 4mM and V2=16 mL
1) 1.0 mM
M1V1=M2V2
V1= M2V2/M1
= 1*16/4
=4 mL
2)250 μM
1 uM= 10^-3 mM
250 *10^-3 mM
M1V1=M2V2
V1= M2V2/M1
= 250*10^-3*16/4=1
V1= 1mL
3) 62.5 μM
1 uM= 10^-3 mM
62.5*10^-3 mM
M1V1=M2V2
V1= M2V2/M1
= 62.5*10^-3*16/4=1/4=0.25
V1= 0.25 mL=250uL ( 1 mL=1000 uL)
4) 15.6 μM
1 uM= 10^-3 mM
15.6*10^-3 mM
M1V1=M2V2
V1= M2V2/M1
= 15.6*10^-3*16/4=0.0624mL
V1=0.0624 *1000=62.4 uL
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