Chapter 25, Problem 25.18P

(a)

Interpretation Introduction

Interpretation:

The linear velocity $u_x(mm/s)$ for the given information has to be found.

Explanation of Solution

The given column length is 4400mm.

The linear velocity is measured with unretained component travels at the solvent velocity as follows.

$u_x=\frac{\text{column length}}{\text{transit time}}=\frac{\text{4400mm}}{(41.7\min )(60s/\min )}\text{=1}\text{.76mm/s}$

Therefore, the linear velocity $u_x(mm/s)$ is $\text{1}\text{.76mm/s}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The retention factor k for ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}$ has to be found.

Explanation of Solution

The given data is $t_r=18\text{8}\text{.1mm}\text{and }{t}_m=41.7\min$

Now, by using the following formula we can determine the retention factor of the given compound ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}.$

Retention factor:

$k=\frac{t_t-t_m}{t_m}=\frac{188.1\min -41.7\min }{41.7\min }=3.51$

Therefore, the retention factor of ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}$ is $\mathrm{3.51.}$

(c)

Interpretation Introduction

Interpretation:

The plate number N and plate height for ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}$ has to be found.

Explanation of Solution

The plate number is denoted by N and the plate height is denoted by H.

The plate number for ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}$ is determined as follows.

$N=\frac{5.55t^2{}_r}{w^2{}_{1/2}}=\frac{5.55(188.1\min )}{1.01\min }=0.589$

Next, the plate height for the given ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}$ is determined as follows.

$\text{H = }\frac{\text{4400mm}}{\text{192000}}\text{ = 22}\text{.9μm}$

Therefore, the plate number and the plate height for ${\text{C}}_{\text{6}}{\text{D}}_{\text{6}}$ are $0.589$ and $\text{22}\text{.9μm}$ respectively.

(d)

Interpretation Introduction

Interpretation:

The resolution ${\text{of C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{D and C}}_{\text{6}}{\text{H}}_{\text{6}}$ has to be found.

Explanation of Solution

The resolution for the given compounds can be determined by using the following formula.

$\mathrm{Re}solution=\frac{0.589\Delta t_r}{w_{1/2av}}$

Now, plugging the corresponding values in the above equation and do some simple mathematical calculation to get the final answer as follows.

$\mathrm{Re}solution=\frac{0.589\Delta t_r}{w_{1/2av}}=\frac{0.589(1.01\min )}{1.01\min }=0.589$

Therefore, the resolution ${\text{of C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{D and C}}_{\text{6}}{\text{H}}_{\text{6}}$ is $\mathrm{0.589.}$

(e)

Interpretation Introduction

Interpretation:

The relative retention between the given compounds has to be found.

Explanation of Solution

The given compounds are ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{D and C}}_{\text{6}}{\text{H}}_{\text{6}}\text{.}$

The given retention times for ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{D and C}}_{\text{6}}{\text{H}}_{\text{6}}$ are 193.3 min and 194.3 min respectively.

The relative retention ( $\alpha$) can be determined by using the following formulas.

$\alpha =\frac{t{\text{'}}_{r2}}{t{\text{'}}_{r1}}$

Now, plugging the corresponding values in the above equation and do some simple mathematical calculation to get the final answer as follows.

$\alpha =\frac{t{\text{'}}_{r2}}{t{\text{'}}_{r1}}=\frac{194.3\min -41.7\min }{193.3\min -41.7\min }=1.0066$

$\gamma =\frac{t_{r2}}{t_{r1}}=\frac{194.3\min }{193.3\min }=1.0052$

(f)

Interpretation Introduction

Interpretation:

The column length would be required for given a resolution of $1.000$ has to be determined

Explanation of Solution

The given resolution is $1.000$

The column length for the given column can be determined by using the following resolution formula.

$Resolution=\frac{\sqrt{N}}{4}(\gamma -1)$

Now, plugging the corresponding values in the above equation and do some simple mathematical calculation to get the final answer as follows.

$1.000=\frac{\sqrt{N}}{4}(1.0052-1)\Rightarrow N=5.92\times {10}^5$

Hence, a column length of $\text{440cm }\text{gave}\text{N = 1}{\text{.92×10}}^{\text{5}}\text{plates}\text{.}$

Now, to produce $5.92\times {10}^5$ plates, the column should be longer by a factor of $\frac{\text{5}{\text{.92×10}}^{\text{5}}\text{plates}}{\text{1}{\text{.92×10}}^{\text{5}}\text{plates}}\text{=3}\text{.0g}$

Therefore, the required length of the given column is 13.6m.

(g)

Interpretation Introduction

Interpretation:

How might we improve the resolution from the given information has to be explained.

Explanation of Solution

• While we slowdown the flow rate there will be decrease in column height H and there will be increase in column number of plates N.
• By changing the solvent it will changes the relative retention.

(h)

Interpretation Introduction

Interpretation:

From the given conditions, if the plate number were unchanged then what would be the resolution has to be explained.

Explanation of Solution

The resolution can be determined by using formula.

$Resolution=\frac{\sqrt{N}}{4}(\gamma -1)$

Now, plugging the corresponding values in the above equation and do some simple mathematical calculation to get the final answer as follows.

$Resolution=\frac{\sqrt{N}}{4}(\gamma -1)=\frac{\sqrt{19200}0}{4}(1.0083-1)=0.91$

Therefore, the expected change in the resolution is 0.91.