A chromatographic procedure separatesof unknown mixture on a column with a length of 40cmand a diameter of 0.85cm . (a) What size column would you use to separate 100mgof the same mixture? (b) If the flow is 0.22mL/minon the small column, what volume flow rate should be used on the large column? (c) The mobile phase occupies 35%of the column volume. Calculate the linear velocity for the small column and the large column.

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 is retained in the mobile phase.

$m_1=4mg,l=40cm,d=0,85cm$

To compare the size of the two columns are

$$\begin{gathered} \frac{m_2}{m_1}={\left(\frac{d_2}{d_1}\right)}^2 \\ \frac{100mg}{4mg}={\left(\frac{d_2}{0.85cm}\right)}^2 \\ 25=\frac{d_2^2}{0.7225c{m}^2} \\ {d}_2^2=18.06c{m}^2 \\ {d}_2=\sqrt{18.06c{m}^2} \\ {d}_2=4.25cm \end{gathered}$$

The size of the column would be 40cmwith a diameter of 4.25cm

If the volume flow is 0.22mL/min on the smaller column, to determine the volume flow rate should be used on the larger column

$$\begin{gathered} \frac{m_2}{m_1}=\frac{u_2}{u_1} \\ \frac{100mg}{4mg}=\frac{u_2}{0.22mL/min} \\ 25\times 0.22mL/min \\ 5.5mL/min=u_2 \end{gathered}$$

To find the volume of a small column as

$$\begin{gathered} V_1=\tau \tau r^2\times l \\ {V}_1=\tau \tau \times {\left(0.85/2cm\right)}^2\times 40cm \\ {V}_1=22.7mL \end{gathered}$$

Volume of a mobile phase as

$$\begin{gathered} V_m=0.35\times 22.7mL \\ {V}_m=7.49mL \end{gathered}$$

To calculate the linear flow

$$\begin{gathered} u_x=\frac{l}{V_m\times u_1} \\ {u}_x=\frac{40cm}{7.94mL\times 0.22mL/min} \\ {u}_x=1.11cm/min \end{gathered}$$