Question

(a) How many moles of analyte are present in a$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{cm}\mathbf{\times }\mathbf{25}\mathbf{\mu m}$solution that occupies 1.0% of the length of a$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{cm}\mathbf{\times }\mathbf{25}\mathbf{\mu m}$capillary?(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,${\mathbf{\mu }}_{\mathbf{mp}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{(}\mathbf{V}\mathbf{\times }\mathbf{s}\mathbf{)}$ and the sample concentration is$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu M}\mathbf{?}$

Short answer

a)n=29.5fmoles of analyte are present in a$10.0-\mathrm{\mu M}$ solution.

b) V oltage required to inject this many moles into a capillary in 4.0 s is${\mathrm{V}}_{\mathrm{v}}=3\times {10}^3\mathrm{V}$

Step-by-step solution

Step 1:Calculate mass and voltage

In this we will calculate the following:

a)the moles of analyte present in$10.0-\mathrm{\mu M}$ 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.

Step 2:Calculate the volume of the sample

a)First we will calculate the volume of the sample

$$\begin{gathered} \mathrm{V}={\mathrm{\pi r}}^2\mathrm{l} \\ \mathrm{V}=\mathrm{\pi }(12.5\times {10}^6\mathrm{m}{)}^2\times (0.006\mathrm{m}) \\ \mathrm{V}=2.95\times {10}^{-12}{\mathrm{m}}^3=2.95\mathrm{nL} \end{gathered}$$

Step 3:Calculate the moles by the following

Next calculate the moles by the following:

$$\begin{gathered} \mathrm{n}=\mathrm{c}\times \mathrm{V} \\ \mathrm{n}=(10\times {10}^{-6}\mathrm{M})\times (2.95\times {10}^{-9}\mathrm{L}) \\ \mathrm{n}=29.5\mathrm{fmol} \end{gathered}$$

Step 4:Convert the concentration to mol/m3

We will convert the concentration to $\mathrm{mol}/{\mathrm{m}}^3$

$10\times {10}^{-6}\mathrm{mol}/{\mathrm{dm}}^3=1\times {10}^{-2}\mathrm{mol}/{\mathrm{m}}^3$

Step 5:Calculate the voltage

Then we will use the following equation in order to calculate the voltage$\left({\mathrm{V}}_{\mathrm{v}}\right)$

$$\begin{gathered} \mathrm{n}={\mathrm{\mu }}_{\mathrm{app}}\left(\frac{\mathrm{E}\times {\mathrm{k}}_{\mathrm{b}}}{{\mathrm{k}}_{\mathrm{s}}}\right){\mathrm{t\pi r}}^2\mathrm{c} \\ \mathrm{n}={\mathrm{\mu }}_{\mathrm{app}}\left(\frac{{\mathrm{V}}_{\mathrm{v}}\times {\mathrm{k}}_{\mathrm{b}}}{{\mathrm{L}}_{\mathrm{t}}\times {\mathrm{k}}_{\mathrm{s}}}\right){\mathrm{t\pi r}}^2\mathrm{c} \end{gathered}$$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{v}}=\frac{\mathrm{n}\times {\mathrm{L}}_{\mathrm{t}}\times {\mathrm{k}}_{\mathrm{s}}/{\mathrm{k}}_{\mathrm{b}}}{{\mathrm{\mu }}_{\mathrm{app}}{\mathrm{t\pi r}}^2\mathrm{c}} \\ {\mathrm{V}}_{\mathrm{v}}=\frac{\left(2.95\times {10}^{-15}\mathrm{mol}\right)\left(0.6\mathrm{m}\right)\left(1/10\right)}{\left(3\times {10}^{-8}{\mathrm{m}}^2/\mathrm{Vs}\right)\left(4\mathrm{s}\right)\mathrm{\pi }\left(12.5\times {10}^{-6}\mathrm{m}\right)\left(1\times {10}^{-2}\mathrm{mol}/{\mathrm{m}}^3\right)} \\ {\mathrm{V}}_{\mathrm{v}}=3\times {10}^3\mathrm{V} \end{gathered}$$