Chapter 26, Problem 26.36P

(a)

Interpretation Introduction

Interpretation:

The number of moles of analyte are present in $\text{10}\text{.0}\text{μM}$ of solution in $\text{1\%}$ occupied $\text{60}\text{cm×25}\text{μM}$ long capillary tube should be calculated.

Concept introduction:

Molarity:

The gram mole of solute that is present in litter volume of solution is known as molarity.

$\text{Molatity}\text{=}\frac{\text{mole}}{\text{volume}}$

Volume of capillary tube:

In the capillary electrophoresis, the volume of capillary tube is is given by below equation,

$\text{V}\text{=π}{\text{r}}^{\text{2}}\text{l}$

Where,

$\text{V}$ is volume of capillary

$\text{r}$ is radius of capillary

$\text{L}$ is total length of the capillary

(b)

Interpretation Introduction

Interpretation:

The required voltage to inject this many moles of sample into a capillary should be calculated.

Concept introduction:

Electro kinetic injection:

In the capillary electrophoresis, the sample is injected by electrical field is known as electro kinetic injection.

In the electro kinetic injection, the injected mole of sample at t time is given by below equation,

$\text{Mole}\text{=}{\text{μ}}_{\text{app}}\left(\text{E}\frac{{\text{k}}_{\text{b}}}{{\text{K}}_{\text{s}}}\right)\text{t}\text{π}{\text{r}}^{\text{2}}\text{C}$

Where,

${\text{μ}}_{\text{app}}$ is apparent mobility

$\text{E}$ is applied electrical field

$\text{r}$ is capillary radius

$\text{C}$ is sample concentration