An open tubular column is 30.01 mlong and has an inner diameter of 0.530mm. It is coated on the inside wall with a layer of stationary phase that is$\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{\mu m}$thick. Unretained solute passes through in 2.16min, whereas a particular solute has a retention time of 17.32min. (a) Find the linear velocity and volume flow rate. (b) Find the retention factor for the solute and the fraction of time spent in the stationary phase.(c) Find the partition coefficient, K 5 cs/cm, for this solute.

The retention factor, also known as the capacity factor (k), in column chromatography is defined as the ratio of time an analyte is retained in the stationary phase to time the analyte spent in the mobile phase.

$$\begin{gathered} \mathrm{I}=30.1\mathrm{m}, \\ {\mathrm{d}}_{\mathrm{outer}}=0.530\mathrm{mm}=530\mathrm{\mu m} \\ {\mathrm{t}}_{\mathrm{m}}=2.16\min , \\ {\mathrm{t}}_{\mathrm{r}}=17.32\min \end{gathered}$$

To calculate the linear velocity and volume flow rate

Linear flow rate as

$u_X=\frac{I}{t_m}$

Substitute the given data values are

$$\begin{gathered} {\mathrm{u}}_{\mathrm{X}}=\frac{30.01\mathrm{m}}{2.16\min } \\ {\mathrm{u}}_{\mathrm{X}}=13.9\mathrm{m}/\min \end{gathered}$$

To calculate the inner diameterof $3.1\mathrm{\mu m}$thick wall of stationary phase

$$\begin{gathered} {\mathrm{d}}_{\mathrm{inner}}={\mathrm{d}}_{\mathrm{outer}}-2\times 3.1\mathrm{\mu m} \\ {\mathrm{d}}_{\mathrm{inner}}=523.8\mathrm{\mu m} \\ \mathrm{r}={\mathrm{d}}_{\mathrm{inner}}/2 \\ \mathrm{r}=261.9\mathrm{\mu m} \end{gathered}$$

To find the volume flow rate:

$$\begin{gathered} V_m=\tau \tau r^{2}I \\ {V}_m=\tau \tau \times {\left(261.9\times {10}^{-4}cm\right)}^2\times 30.1\times {10}^{2}cm \\ {V}_m=6.49c{m}^3 \\ {V}_m=6.49mL \\ {u}_v=\frac{V_m}{t_m} \\ {u}_v=\frac{6.49mL}{2.16min} \\ {u}_v=3mL/min \end{gathered}$$

To calculate the retention factor

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ \mathrm{k}=\frac{17.32\min -2.16\min }{2.16\min } \\ \mathrm{k}=7.02 \\ \frac{{\mathrm{t}}_{\mathrm{s}}}{{\mathrm{t}}_{\mathrm{s}}+{\mathrm{t}}_{\mathrm{m}}}=\frac{\mathrm{k}\times {\mathrm{t}}_{\mathrm{m}}}{\mathrm{k}\times {\mathrm{t}}_{\mathrm{m}}+{\mathrm{t}}_{\mathrm{m}}} \\ \frac{{\mathrm{t}}_{\mathrm{S}}}{{\mathrm{t}}_{\mathrm{S}}+{\mathrm{t}}_{\mathrm{m}}}=\frac{\mathrm{k}}{\mathrm{k}+1} \\ \frac{{\mathrm{t}}_{\mathrm{S}}}{{\mathrm{t}}_{\mathrm{S}}+{\mathrm{t}}_{\mathrm{m}}}=\frac{7.02}{7.02+1} \\ \frac{{\mathrm{t}}_{\mathrm{S}}}{{\mathrm{t}}_{\mathrm{S}}+{\mathrm{t}}_{\mathrm{m}}}=0.875 \end{gathered}$$

To find the volume of a coat inside the tube is

$$\begin{gathered} {\mathrm{V}}_2=2\mathrm{\tau \tau r}\times \mathrm{thickness}\times \mathrm{I} \\ {\mathrm{V}}_{\mathrm{S}}=2\mathrm{\tau \tau }\left(261.9\times {10}^{-4}\mathrm{cm}+1.55\times {10}^{-4}\mathrm{cm}\right)\times \left(3.1\times {10}^{-4}\mathrm{cm}\right)\times \left(30.1\times {10}^2\mathrm{cm}\right) \\ {\mathrm{V}}_{\mathrm{S}}=0.154{\mathrm{cm}}^3 \\ {\mathrm{V}}_{\mathrm{S}}=0.154\mathrm{mL} \end{gathered}$$

To find the Partition coefficient

$$\begin{gathered} \mathrm{k}=\mathrm{K}\frac{{\mathrm{V}}_{\mathrm{S}}}{{\mathrm{V}}_{\mathrm{m}}} \\ \mathrm{K}=\mathrm{k}\times \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}} \\ \mathrm{K}=7.02\times \frac{6.49\mathrm{mL}}{0.154\mathrm{mL}} \\ \mathrm{K}=295.8 \end{gathered}$$