Chapter 23, Problem 23.51P

(a)

Interpretation Introduction

Interpretation:

The value of adjusted retention time and retention factors for pentafluorobenzene and benzene has to be calculated.

Concept introduction:

Retention factor:

The retention factor $\left(\text{k}\right)$ is given by

$\begin{array}{l}\text{k}\text{=}\frac{{\text{t}}_{\text{r}}\text{-}{\text{t}}_{\text{m}}}{{\text{t}}_{\text{m}}}\\ \text{Where,}\\ \text{k}\text{-}\text{retention}\text{factor}\\ {\text{t}}_{\text{r}}\text{-}\text{time}\text{required}\text{to}\text{elute}\text{the}\text{peak}\\ {\text{t}}_{\text{m}}\text{-}\text{time}\text{required}\text{for}\text{mobile}\text{phase}\text{to}\text{pass}\text{through}\text{the}\text{column}\text{.}\end{array}$

Answer to Problem 23.51P

• The value of adjusted retention time for pentafluorobenzene is $\text{11}\text{.92}\text{min}\text{.}$
• The value of retention factor for pentafluorobenzene is $\text{11}\text{.25}\text{.}$
• The value of adjusted retention time for benzene is $\text{12}\text{.14}\text{min}\text{.}$
• The value of retention factor for benzene is $\text{11}\text{.45}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Elution}\text{time}\text{for}\text{unretained}\text{solute}\text{is}\text{1}\text{.06}\text{min}\text{.}\\ \text{Length}\text{of}\text{the}\text{column}\text{is}\text{30}\text{.0}\text{m}\text{and}\text{diameter}\text{is}\text{0}\text{.530}\text{mm}\text{.}\\ \text{Thickness}\text{of}\text{the}\text{stationary}\text{phase}\text{on}\text{the}\text{inner}\text{wall}\text{is}\text{3}\text{.0}\text{µm}\text{.}\end{array}$

Calculate the value of adjusted retention time and retention factor for pentafluorobenzene

$\begin{array}{l}{\text{t}}_{\text{r}}^{\text{'}}\text{=}\text{12}\text{.98}\text{-}\text{1}\text{.06}\text{=}\text{11}\text{.92}\text{min}\text{.}\\ \\ \text{k}\text{=}\frac{\text{11}\text{.92}}{\text{1}\text{.06}}\text{=}\text{11}\text{.25}\end{array}$

Calculate the value of adjusted retention time and retention factor for benzene

$\begin{array}{l}{\text{t}}_{\text{r}}^{\text{'}}\text{=}\text{13}\text{.20}\text{-}\text{1}\text{.06}\text{=}\text{12}\text{.14}\text{min}\text{.}\\ \\ \text{k}\text{=}\frac{\text{12}\text{.14}}{\text{1}\text{.06}}\text{=}\text{11}\text{.45}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of relative retention $\left(\text{α}\right)$ has to be calculated.

Answer to Problem 23.51P

• The value of relative retention $\left(\text{α}\right)$ is $\text{1}\text{.018}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}{\text{t}}_{\text{r}}^{\text{'}}\text{value}\text{of}\text{benzene}\text{is}\text{12}\text{.14}\text{min}\text{.}\\ {\text{t}}_{\text{r}}^{\text{'}}\text{value}\text{of}\text{pentafluorobenzene}\text{is}\text{11}\text{.92}\text{min}\text{.}\end{array}$

Calculate the value of relative retention $\left(\text{α}\right)$

$\text{α}\text{=}\frac{\text{12}\text{.14}}{\text{11}\text{.92}}\text{=}\text{1}\text{.018}\text{.}$

The value of relative retention $\left(\text{α}\right)$ is $\text{1}\text{.018}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The number of plates $\left(\text{N}\right)$ and plate height has to be calculated.

Concept introduction:

Plates of column:

The relationship between $\text{N}\text{and}{\text{w}}_{\text{1/2}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{5}\text{.55}{\text{t}}_{\text{r}}{}^{\text{2}}}{{\text{w}}_{\text{1/2}}{}^{\text{2}}}\\ \text{Where,}\\ \text{N}\text{-}\text{number}\text{of}\text{plates}\text{on}\text{column}\\ {\text{t}}_{\text{r}}\text{-}\text{retention}\text{time}\\ {\text{w}}_{\text{1/2}}\text{-}\text{half-width}\end{array}$

Answer to Problem 23.51P

• The number of plates for pentafluorobenzene is $\text{6}\text{.08}\text{×}{\text{10}}^{\text{4}}\text{.}$
• The plate height for pentafluorobenzene is $\text{0}\text{.493}\text{mm}\text{.}$
• The number of plates for benzene is $\text{6}\text{.60}\text{×}{\text{10}}^{\text{4}}\text{.}$
• The plate height for benzene is $\text{0}\text{.455}\text{mm}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}{\text{w}}_{\text{1/2}}\left({\text{C}}_{\text{6}}{\text{HF}}_{\text{5}}\right)\text{=}\text{0}\text{.124}\text{min}\text{.}\\ {\text{w}}_{\text{1/2}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)\text{=}\text{0}\text{.121}\text{min}\text{.}\\ \text{Column}\text{length}\text{=}\text{30}\text{.0}\text{m}\text{.}\end{array}$

Calculate the number of plates and plate height for ${\text{C}}_{\text{6}}{\text{HF}}_{\text{5}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{5}\text{.55}{\text{t}}_{\text{r}}^{\text{2}}}{{\text{w}}_{\text{1/2}}^{\text{2}}}\text{=}\frac{\text{5}\text{.55}{\left(\text{12}\text{.98}\right)}^{\text{2}}}{{\left(\text{0}\text{.124}\right)}^{\text{2}}}\text{=}\text{6}\text{.08}\text{×}{\text{10}}^{\text{4}}\text{plates}\text{.}\\ \\ \text{Plate}\text{height}\left(\text{H}\right)\text{=}\frac{\text{L}}{\text{N}}\text{=}\frac{\text{30}\text{.0}\text{m}}{\text{6}\text{.08}\text{×}{\text{10}}^{\text{4}}}\text{=}\text{0}\text{.493}\text{mm}\text{.}\end{array}$

Calculate the number of plates and plate height for ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{5}\text{.55}{\text{t}}_{\text{r}}^{\text{2}}}{{\text{w}}_{\text{1/2}}^{\text{2}}}\text{=}\frac{\text{5}\text{.55}{\left(\text{13}\text{.20}\right)}^{\text{2}}}{{\left(\text{0}\text{.121}\right)}^{\text{2}}}\text{=}\text{6}\text{.60}\text{×}{\text{10}}^{\text{4}}\text{plates}\text{.}\\ \\ \text{Plate}\text{height}\left(\text{H}\right)\text{=}\frac{\text{L}}{\text{N}}\text{=}\frac{\text{30}\text{.0}\text{m}}{\text{6}\text{.60}\text{×}{\text{10}}^{\text{4}}}\text{=}\text{0}\text{.455}\text{mm}\text{.}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The number of plates has to be calculated by using the value of width.

Concept introduction:

Plates of column:

The relationship between $\text{N}\text{and}\text{w}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{16}{\text{t}}_{\text{r}}{}^{\text{2}}}{{\text{w}}^{\text{2}}}\\ \text{Where,}\\ \text{N}\text{-}\text{number}\text{of}\text{plates}\text{on}\text{column}\\ {\text{t}}_{\text{r}}\text{-}\text{retention}\text{time}\\ \text{w}\text{-}\text{half-width}\end{array}$

Answer to Problem 23.51P

• The number of plates for pentafluorobenzene is $5.57\text{×}{\text{10}}^{\text{4}}\text{.}$
• The number of plates for benzene is $4.88\text{×}{\text{10}}^{\text{4}}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{w}\text{for}\left({\text{C}}_{\text{6}}{\text{HF}}_{\text{5}}\right)\text{=}\text{0}\text{.220}\text{min}\text{.}\\ \text{w}\text{for}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)\text{=}\text{0}\text{.239}\text{min}\text{.}\end{array}$

Calculate the number of plates for ${\text{C}}_{\text{6}}{\text{HF}}_{\text{5}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{16}{\text{t}}_{\text{r}}^{\text{2}}}{{\text{w}}^{\text{2}}}\\ \text{=}\frac{\text{5}\text{.55}{\left(\text{12}\text{.98}\right)}^{\text{2}}}{{\left(\text{0}\text{.220}\right)}^{\text{2}}}\text{=}\text{5}\text{.57}\text{×}{\text{10}}^{\text{4}}\text{plates}\text{.}\end{array}$

Calculate the number of plates for ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{16}{\text{t}}_{\text{r}}^{\text{2}}}{{\text{w}}^{\text{2}}}\\ \text{=}\frac{\text{16}{\left(\text{13}\text{.20}\right)}^{\text{2}}}{{\left(\text{0}\text{.239}\right)}^{\text{2}}}\text{=}\text{4}\text{.88}\text{×}{\text{10}}^{\text{4}}\text{plates}\text{.}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The resolution between the given two peaks has to be calculated.

Answer to Problem 23.51P

• The resolution between the given two peaks is $\text{0}\text{.96}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Value}\text{of}{\text{Δt}}_{\text{r}}\text{=}\text{13}\text{.20}\text{-}\text{12}\text{.98}\text{=}\text{0}\text{.22}\\ \text{Value}\text{of}{\text{w}}_{\text{av}}\text{is}\text{0}\text{.229}\text{.}\end{array}$

Calculate the resolution between the given two peaks

$\begin{array}{l}\text{Resolution}\text{=}\frac{{\text{Δt}}_{\text{r}}}{{\text{w}}_{\text{av}}}\\ \text{=}\frac{\text{13}\text{.20}\text{-}\text{12}\text{.98}}{\text{0}\text{.229}}\text{=}\text{0}\text{.96}\end{array}$

The resolution between the given two peaks is $\text{0}\text{.96}\text{.}$

(f)

Interpretation Introduction

Interpretation:

The numerical value of resolution has to be calculated.

Answer to Problem 23.51P

• The numerical value of resolution is $\text{0}\text{.97}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}{\text{N}}_{\text{1}}\text{=}\text{5}\text{.57}\text{×}{\text{10}}^{\text{4}}\text{plates}\text{.}\\ {\text{N}}_{\text{2}}\text{=}\text{4}\text{.88}\text{×}{\text{10}}^{\text{4}}\text{plates}\text{.}\end{array}$

Calculate the number of plates

$\begin{array}{l}\text{N}\text{=}\sqrt{{\text{N}}_{\text{1}}{\text{N}}_{\text{2}}}\text{=}\sqrt{\left(\text{5}\text{.57}\text{×}{\text{10}}^{\text{4}}\right)\left(\text{4}\text{.88}\text{×}{\text{10}}^{\text{4}}\right)}\\ \\ \text{=}\sqrt{\text{2718160000}}\\ \\ \text{=}\text{5}\text{.21}\text{×}{\text{10}}^{\text{4}}\text{plates}\text{.}\end{array}$

Calculate the resolution

$\begin{array}{l}\text{Resolution}\text{=}\frac{\sqrt{\text{N}}}{\text{4}}\left(\text{γ}\text{-}\text{1}\right)\\ \\ \text{=}\frac{\sqrt{\text{5}\text{.21}\text{×}{\text{10}}^{\text{4}}}}{\text{4}}\left(\text{1}\text{.017}\text{-}\text{1}\right)\text{=}\text{0}\text{.97}\text{.}\end{array}$

The resolution is $\text{0}\text{.97}\text{.}$