91Ó°ÊÓ

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.

$$\begin{gathered} \mathrm{I}=30.1\mathrm{m}, \\ {\mathrm{d}}_{\mathrm{outer}}=0.530\mathrm{mm}=530\mathrm{μ³¾} \\ {\mathrm{t}}_{\mathrm{m}}=2.16\min , \\ {\mathrm{t}}_{\mathrm{r}}=17.32\min \end{gathered}$$

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 $3.1\mathrm{μ³¾}$thick wall of stationary phase

$$\begin{gathered} {\mathrm{d}}_{\mathrm{inner}}={\mathrm{d}}_{\mathrm{outer}}-2×3.1\mathrm{μ³¾} \\ {\mathrm{d}}_{\mathrm{inner}}=523.8\mathrm{μ³¾} \\ \mathrm{r}={\mathrm{d}}_{\mathrm{inner}}/2 \\ \mathrm{r}=261.9\mathrm{μ³¾} \end{gathered}$$

To find the volume flow rate:

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

To calculate the retention factor

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

To find the volume of a coat inside the tube is

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

To find the Partition coefficient

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