91Ó°ÊÓ

Interpretation:

The separation of Blue Dextran has 2000 in gel filtration column using should be explained.

Concept introduction:

Gel filtration in open columns:

In the gel filtration column, the pore size of the stationary phase is chosen depends on the molecular mass of the molecules to be separated.

The fractionation range for globular proteins (Da) are used as gel in gel filtration columns and more size of gel are available based on it is size so the stationary phase size (Da) is depends on molar mass of the compound, which is going to separate.

The (Da) value of Sephadex-G 50 is 1500-30000 but Blue Dextran has 2000 molar mass so it is not eluted from this column.

Explanation of Solution

To explain the separation of Blue Dextran 2000 in gel filtration column (Sephadex- G 50 is 1500-30000 ). In the gel-filtration column, the (Da) value of Sephadex (stationary phase) is directly proportional to the molar mass of compound, which is going to separate.

Therefore, the Da value of given stationary phase Sephadex- G 50 and molar mass of the Blue Dextran is 2000 so it does not elute from this column.

(b) Interpretation Introduction

Interpretation:

The retention volume of a molecule in gel-filtration column should be determined.

Concept introduction:

The volume of column is given by the mathematical expression it is,

${\mathrm{V}}_{\mathrm{t}}={\mathrm{Ï€°ù}}^2\mathrm{l}$

Where,

V_tis volume of column

$\mathrm{Ï€}$is 3.14

R is radius of the column

L is length of the column

The Elution Equation :

${\mathrm{K}}_{\mathrm{av}}=\frac{{\mathrm{V}}_{\mathrm{r}}-{\mathrm{V}}_0}{{\mathrm{V}}_{\mathrm{m}}-{\mathrm{V}}_0}$

Retention volume of NaCl is 84.1 mL

Explanation of Solution

To determine the retention volume of a molecule in gel-filtration column should be determined.

Given,

The initial volume is 36.4mL

The total volume of mobile phase is 36.4 mL

Therefore, retention volume of NaCl is,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{uv}}=\frac{{\mathrm{V}}_{\mathrm{r}}-36.4}{109.8-36.4} \\ 0.65=\frac{{\mathrm{V}}_{\mathrm{r}}-36.4}{109.8-36.4} \\ {\mathrm{V}}_{\mathrm{r}}-36.4=0.65×109.8-36.4 \\ {\mathrm{V}}_{\mathrm{r}}=84.1\mathrm{mL} \end{gathered}$$

The given volume and ${\mathrm{K}}_{\mathrm{av}}$ are plugged in above equation to give a retention volume of NaCl given column Retention volume of NaCl is 84.1mL