Given a stock solution of compound Y with a concentration of 0.15 M, calculate t

Question

Given a stock solution of compound Y with a concentration of 0.15 M, calculate the oncentrations of the 5 prepared solutions assuming you perform these dilutions exactly as require n the procedure of this lab. For example, 0.20 mL into 25.00 mL; 040 mL. into 25.00 mL; et.. If you are told that the purity of compound Z is 75% and the reported value for its molar xtinction coefficient is 23000 M-1cm1 at 611 nm, what should you measure as the of this impu ompound Z? . Why is it important to insert the cuvette using the same orientation for all spectrophotometer measurements? 119

Explanation / Answer

1) different concentrations:

M1= initial concentration    V1= initial volume

M2= final concentraiton      V2= final volume

M1V1 = M2V2

a) M1 =0.15 M   V1 = 0.2mL

M2 = ? V2 = 25mL

M2 = M1V1/ V2 = 0.15 X 0.2 / 25 = 0.0012 M

b)M1 =0.15 M   V1 = 0.4mL

M2 = ? V2 = 25mL

M2 = M1V1/ V2 = 0.15 X 0.4 / 25 = 0.0024 M

c) M1 =0.15 M   V1 = 0.6mL

M2 = ? V2 = 25mL

M2 = M1V1/ V2 = 0.15 X 0.6 / 25 = 0.0036 M

d) M1 =0.15 M   V1 = 0.8mL

M2 = ? V2 = 25mL

M2 = M1V1/ V2 = 0.15 X 0.8 / 25 = 0.0048 M

e) M1 =0.15 M   V1 = 1mL

M2 = ? V2 = 25mL

M2 = M1V1/ V2 = 0.15 X 1 / 25 = 0.0060 M

2) according to Beer law

Absorbance = concentration X extinction coeffecient X path length

so the extinction coeffecient will be

Extinction coeffecient = 23000 / 0.75 = 30666.67

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