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(a) How many moles of analyte are present in a60.0cm×25μ³¾solution that occupies 1.0% of the length of a60.0cm×25μ³¾capillary?(b) What voltage is required to inject this many moles into a capillary in 4.0 s if the sample has 1/10 of the conductivity of background electrolyte,μmp=3.0×10-8m2/(V×s) and the sample concentration is10.0μ²Ñ?

Short Answer

Expert verified

a)n=29.5fmoles of analyte are present in a10.0-μ²Ñ solution.

b) V oltage required to inject this many moles into a capillary in 4.0 s isVv=3×103V

Step by step solution

01

Step 1:Calculate mass and voltage

In this we will calculate the following:

a)the moles of analyte present in10.0-μ²Ñ solution that occupies 1% of the length of a 60 cm*25capillary.

b)the voltage required in order to inject that many moles into a capillary in 4 s.

02

Step 2:Calculate the volume of the sample

a)First we will calculate the volume of the sample

V=Ï€°ù2lV=Ï€(12.5×106m)2×(0.006m)V=2.95×10-12m3=2.95nL

03

Step 3:Calculate the moles by the following

Next calculate the moles by the following:

n=c×Vn=(10×10-6M)×(2.95×10-9L)n=29.5fmol

04

Step 4:Convert the concentration to mol/m3

We will convert the concentration to mol/m3

10×10-6mol/dm3=1×10-2mol/m3

05

Step 5:Calculate the voltage

Then we will use the following equation in order to calculate the voltageVv

n=μappE×kbkstÏ€°ù2cn=μappVv×kbLt×kstÏ€°ù2c

Vv=n×Lt×ks/kbμapptÏ€°ù2cVv=2.95×10-15mol0.6m1/103×10-8m2/Vs4sÏ€12.5×10-6m1×10-2mol/m3Vv=3×103V

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