91影视

Compute the adjusted retention time$({t_r}^{\text{'}})$by applying the formula below:

${{\mathrm{t}}_{\mathrm{r}}}^{\text{'}}={\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}$

8.4min-3.7 min

4.7 min

The retention factor is the time required to elute that peak minusthe time from mobile phase to pass through the column, expressed in multiples of${\mathrm{t}}_{\mathrm{m}}$. Calculate the retention factor using the formula below:

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ =\frac{8.4\min -3.7\min }{3.7\min } \end{gathered}$$

GIVEN:

- Diameter of column =$0.32\mathrm{mm}$

- Thickness of stationary phase =$1.0\mathrm{渭尘}$

Find: phase ratiorole="math" localid="1654785803260" $尾$

The phase ratio role="math" localid="1654785811025" $尾$is the volume of the mobile phase divided by the volume of stationary phase. It can be deduced to the following expression:

$\mathrm{尾}=\frac{\mathrm{r}}{2{\mathrm{d}}_{\mathrm{f}}}$

where:

$$\begin{gathered} -\mathrm{r}=\mathrm{column}\mathrm{ratio} \\ -{\mathrm{d}}_{\mathrm{f}}=\mathrm{stationary}\mathrm{fase}\mathrm{film}\mathrm{thickness} \end{gathered}$$

Solve for the phase ratio:

$\mathrm{尾}=\frac{0.32/2\mathrm{mm}}{2\left(1.0\mathrm{渭尘}\right)\left(\begin{array}{c}1\mathrm{mm}\\ 1000\mathrm{渭尘}\end{array}\right)}=80$

Find: partition coefficient$\left(\mathrm{K}\right)$of the analyte

The retention factor $\left(\mathrm{k}\right)$is related to the partition coefficient role="math" localid="1654783983446" $\left(\mathrm{K}\right)$and phase ratio $\left(\mathrm{尾}\right)$, based on the following expression:

$k=\frac{K}{尾}$

Thus, the partition coefficient $\left(K\right)$can be calculated by rearranging the expression:

$\mathrm{K}=\mathrm{k}脳\mathrm{尾}$

$$\begin{gathered} =1.27脳80 \\ =101.6 \end{gathered}$$

Given:

- Diameter of column $=0.32\mathrm{mm}$

- Thickness of stationary phase$=0.5\mathrm{渭尘}$

Find: retention time$({\mathrm{t}}_{\mathrm{r}})$of the same stationary phase and at the same temperature

Based on the formula of phase ratio role="math" localid="1654785312894" $\left(尾\right)$, decreasing the thickness of the stationary phase, increases the role="math" localid="1654785292964" $尾$.

Calculating the new phase ratio role="math" localid="1654784786511" $\left({尾}^{路}\right)$:

role="math" localid="1654785571991" ${尾}^{\text{'}}=\frac{0.32/2mm}{2\left[0.5渭m\right]\left(\begin{array}{c}1mm\\ 1000渭m\end{array}\right)}=160$

$\mathrm{The}\mathrm{new}\mathrm{phase}\mathrm{ratio}\left({尾}^{\text{'}}\right)\mathrm{is}2\mathrm{times}\mathrm{the}\mathrm{original}\mathrm{value}.$

${尾}^{\text{'}}=2尾$

$$\begin{gathered} \mathrm{Thus},\mathrm{the}\mathrm{retention}\mathrm{time}\left({\mathrm{t}}_{\mathrm{m}}\right)\mathrm{is}\mathrm{expected}\mathrm{to}\mathrm{decrease}. \\ \mathrm{Then},\mathrm{compute}\mathrm{for}\mathrm{the}\mathrm{new}\mathrm{retention}\mathrm{factor}\left({\mathrm{k}}^{\text{'}}\right): \end{gathered}$$

${\mathrm{k}}^{\text{'}}=\frac{\mathrm{K}}{{尾}^{\mathit{\text{'}}}}$

thus,

${\mathrm{k}}^{\text{'}}=\frac{\mathrm{K}}{2\mathrm{尾}}=\frac{1}{2}\left(1.27\right)=0.635$

Lastly, compute for the retention time $\left(t_r\right)$:

$$\begin{gathered} {\mathrm{k}}^{\text{'}}=0.635=\frac{{\mathrm{t}}_{\mathrm{r}}-3.7}{3.7} \\ {\mathrm{t}}_{\mathrm{r}}=6.05\min \end{gathered}$$

$$\begin{gathered} a)k=1.27 \\ b)尾=80 \\ c)K=101.6 \\ D)t_r=6.05min \end{gathered}$$