An open tubular column has an inner diameter of $\mathbf{207}\mathbf{\mu m}$and the thickness of the stationary phase on the inner wall is $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{\mu m}$. Unretained solute passes through in 63s and a particular solute emerges in 433s . Find the partition coefficient for this solute and find the fraction of time spent in the stationary phase.

In this task, we have a tubular column with an inner diameter of $207\mathrm{\mu m}$and a stationary phase thickness of $0.50\mathrm{\mu m}$on the inner wall. We'll calculate the partition coefficient for this solute and the fraction of time spent in the stationary phase because the unretained liquid passes through in 63s and a specific solute emerges in 433s .

It is given

Mobile phase

$$\begin{gathered} {\mathrm{d}}_{\mathrm{inner}}=207\mathrm{\mu m}-2\mathrm{thickness}=206\mathrm{\mu m} \\ {\mathrm{r}}_{\mathrm{inner}}={\mathrm{d}}_{\mathrm{inner}}/2=103\mathrm{\mu m} \end{gathered}$$

Stationary phase

$$\begin{gathered} {\text{d}}_{\mathrm{outer}}={\mathrm{d}}_{\mathrm{inner}}+\mathrm{thickness}=206.5\mathrm{\mu m} \\ {r}_{outer}=d_{outer}/2=103.25\mu m \\ {t}_m=63s \\ {t}_r=433s \end{gathered}$$

First we will calculate the retention for this solute:

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ \mathrm{k}=\frac{433\mathrm{s}-63\mathrm{s}}{63\mathrm{s}} \\ \mathrm{k}=5.87 \end{gathered}$$

We know the partition coefficient equation is as follows:

$\mathrm{K}=\mathrm{k}\times \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}$

we know the value of , thus we'll figure out the value of $\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}$

$\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{{\mathrm{\pi r}}^2\times \mathrm{I}}{2\mathrm{\pi r}\times \mathrm{thickness}\times \mathrm{I}}$

Remove the values that are repeated presently.

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{\cancel{\mathrm{\pi }}{\mathrm{r}}^2\times \cancel{\mathrm{i}}}{2\cancel{\mathrm{\pi }}\mathrm{r}\times \mathrm{thickness}\times \cancel{\mathrm{i}}} \\ \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{{\mathrm{r}}_{\mathrm{inner}}^2}{2{\mathrm{r}}_{\mathrm{outer}}\times \mathrm{thickness}} \\ \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{\left(103\mathrm{\mu m}\right)}{2\times 103.25\mathrm{\mu m}\times 0.5\mathrm{\mu m}} \\ \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=102.8 \end{gathered}$$

The partition coefficient can then be calculated as follows:

$$\begin{gathered} \mathrm{K}=\mathrm{k}\times \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}} \\ \mathrm{K}=5.87\times 102.8 \\ \mathrm{K}=603 \end{gathered}$$

We'll also figure out how much time we spent in the stationary phase:

$$\begin{gathered} \mathrm{Fraction}=\frac{{\mathrm{t}}_{\mathrm{s}}}{{\mathrm{t}}_{\mathrm{s}}+{\mathrm{t}}_{\mathrm{m}}} \\ \mathrm{Fraction}=\frac{\mathrm{k}\times {\mathrm{t}}_{\mathrm{m}}}{\mathrm{k}\times {\mathrm{t}}_{\mathrm{m}}+{\mathrm{t}}_{\mathrm{m}}} \\ \mathrm{Fraction}=\frac{\mathrm{k}}{\mathrm{k}+1} \\ \mathrm{Fraction}=\frac{5.87}{5.87+1} \\ \mathrm{Fraction}=0.854 \end{gathered}$$