91影视

Part (a)

First, we calculate the volume at which solvent front appears:

Then we calculated${\mathrm{t}}_{\mathrm{m}}$:

${\mathrm{t}}_{\mathrm{m}}=\frac{\mathrm{L}.{\mathrm{d}}_{\mathrm{c}}^2}{2.\mathrm{F}}$

Or:

${\mathrm{t}}_{\mathrm{m}}=\frac{\mathrm{Vm}}{\mathrm{F}}$

For column A:

$$\begin{gathered} {\mathrm{t}}_{\mathrm{m}}=\frac{0.529{\mathrm{cm}}^3}{1.4{\mathrm{cm}}^3/\min } \\ {\mathrm{t}}_{\mathrm{m}}=0.378\min \end{gathered}$$

For column B:

$$\begin{gathered} {\mathrm{t}}_{\mathrm{m}}=\frac{0.529{\mathrm{cm}}^3}{2{\mathrm{cm}}^3/\min } \\ {\mathrm{t}}_{\mathrm{m}}=0.265\min \end{gathered}$$

Part (b)

The morphine-3-$\mathrm{尾}$-D-glucuronide would elute before morphine because it is more polar (it has more hydroxy groups and one carboxylic acid). So, because the reversed phase column is nonpolar the polar compound would less be retained.

Part (c)

The retention factor can be calculated using the formula:

$\mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}}$

For morphine-3-$\mathrm{尾}$-D-glucuronide, the retention factor is:

$$\begin{gathered} {\mathrm{k}}_1=\frac{\left(1.5-0.65\right)\min }{0.65\min } \\ {\mathrm{k}}_1=1.31 \end{gathered}$$

For morphine, the retention factor is:

$$\begin{gathered} {\mathrm{k}}_2=\frac{\left(2.8-0.65\right)\min }{0.65\min } \\ {\mathrm{k}}_2=3.31 \end{gathered}$$

Part (d)

Column B has bare silica which is polar. Morphine should be retained less than morphine-3-$\mathrm{尾}$-D-glucuronide because it is less polar. The gradient goes from high to low acetonitrile volume fraction because it tends to increase concentration of water, so that the polarity can be increased. Because of that, the more polar compound can be removed.

Part (e)

The average retention factor is calculated using the equation:

${\mathrm{k}}^{*}=\frac{{\mathrm{t}}_{\mathrm{G}}.\mathrm{F}}{鈭�\mathrm{筛}.{\mathrm{V}}_{\mathrm{m}}.\mathrm{S}}$

First, we need to calculate the volume of mobile phase in the column:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{m}}=\mathrm{F}.{\mathrm{t}}_{\mathrm{m}} \\ {\mathrm{V}}_{\mathrm{m}}=2\mathrm{m}\mathrm{L}/\min .0.5\min \\ {\mathrm{V}}_{\mathrm{m}}=1\mathrm{mL} \end{gathered}$$

The ${\mathrm{k}}^{*}$is:

$$\begin{gathered} {\mathrm{k}}^{*}=\frac{5\min .2\mathrm{mL}/\min }{0.4.1\mathrm{mL}.4} \\ {\mathrm{k}}^{*}=6.25 \end{gathered}$$

Here, the final result of part(a), part(b), part(c), part(d), part(e) is

$$\begin{gathered} a){\mathrm{V}}_{\mathrm{m}}=0.529{\mathrm{cm}}^3{\mathrm{t}}_{\mathrm{m}}=0.378\min {\mathrm{t}}_{\mathrm{m}}=0.265\min \\ b)Thepolarcompoundwouldlessberetained \\ c){\mathrm{k}}_1=1.31{\mathrm{k}}_2=3.31 \\ d)Themorepolarcompoundcanberemoved. \\ e){\mathrm{k}}^{*}=6.25 \end{gathered}$$