Question

How many moles of analyte are present in 10.0 μM solution that occupies 1.0% of the length of a 60.0 cm × 25.0 μm capillary?

Answer 1

Answer:

Given data :

Analyte Present = 10.0μM

Solution that occupies = 1.0%

length = 60.0cm X 25.0μm

Capillary = 4.0 seconds

μapp. = 3.0 X 1$0^{-8}$ $m^{2}$ /(V.S)

a) The total a mount of injected analyte is the resultant of electroinjection and EOF  and can e calculated from the equation.

N(A) = CA,A (μeof + ૪ μep) Eot

Mobility of analyse A,૪ is the ratio of the resistivity of the sample to BGS, ''E0'' -> is the electric field strength.

                           ''t'' -> is the time of injection.

So we have to calculate N (A)

Molarity = 10 X 10^-6

Volume = (1/100) X 25 X 10^-6) X (60/100)

             = 15 X 10^8

So moles = N(A) = 15 X 10^13

                 N(A) = 29.5 mol    

Answer 2

Answer:

29.5

D option is correct

None of the above