91Ó°ÊÓ

a. By examining Figure 24-15, the retention time$\left({\mathrm{t}}_{\mathrm{r}}\right)$of peak 11 is approximately

49.5 min .It is given in the problem that thelocalid="1655021317277" $\left({\mathrm{t}}_{\mathrm{m}}\right)$is 6.7 min. Thus, we can calculate the retention factorlocalid="1655021325362" $\left(\mathrm{k}\right)$using the Equation below:

localid="1655021329461" $\mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}}$

Substitute the given values to the equation:

localid="1655021363115" $\mathrm{k}=\frac{49.5\min -6.7\min }{6.7\min }=6.39$

b. Perform approximations by measuring the length of a known retention time in the

graph and solve for the unknown values (w) by measuring the length and applying

ratio and proportion.

For example, the length of a retention time of 2.5 min 10mm(= x in the plot). The

width at the base (w) is 3mm, thus the corresponding retention time is:

$3\mathrm{mm}×\frac{2.5\min }{10\mathrm{mm}}=0.75$

The width at the base of the peak 11 can be approximated to 0.75 min.

Apply Equation 23-30 to compute for the number of theoretical plates (N):

$\mathrm{N}=\frac{16{\mathrm{t}}_{\mathrm{r}}^2}{{\mathrm{w}}^2}=\frac{16{(49.5\min )}^2}{{\left(0.75\min \right)}^2}=6.97×{10}^4$

c. By examining Figure 24-15, the retention time$\left({\mathrm{t}}_{\mathrm{r}}^{\text{'}}\right)$of peak 16 is approximately

62.75 min while that of peak 17 is around 63.5min .

Computing for the$∆{\mathrm{t}}_{\mathrm{r}}$:

$∆{\mathrm{t}}_{\mathrm{r}}=63.5\min -62.75\min =0.75\min$Step 7

The width of the bases of peak 16 and peak 17 are almost equal and

can be approximated to 0.5 min. Thus, the average width of the two peaks $\left({\mathrm{w}}_{\mathrm{av}}\right)$is

0.5 min. Calculating for the resolution between peak 16 and peak 17:

$$\begin{gathered} \mathrm{Resolution}=\frac{∆{\mathrm{t}}_{\mathrm{r}}}{{\mathrm{w}}_{\mathrm{av}}} \\ =\frac{0.75\min }{0.5\min }=1.5 \end{gathered}$$