Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=-\log T$

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8\%$.

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=-\log T$

Where,

• $A$ is the absorbance of the solution.
• $T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

${10}^{-A}={10}^{\left(\log \left(T\%/100\right)\right)}$

Simplify the expression for percent transmittance.

$T\%=\left({10}^{-A}\right)100$   ...... (1)

The value for $A$ is $0.042$.

Substitute the value in above equation.

$\begin{array}{c}T\%=\left({10}^{-A}\right)100\\ T\%=\left({10}^{\left(-0.042\right)}\right)100\\ =\left(0.908\right)100\\ =90.8\%\end{array}$

The percent transmittance for the absorbance of $0.042$ is $90.8\%$.

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=-\log T$

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1\%$.

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=-\log T$

Where,

• $A$ is the absorbance of the solution.
• $T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

${10}^{-A}={10}^{\left(\log \left(T\%/100\right)\right)}$

Simplify the expression for percent transmittance.

$T\%=\left({10}^{-A}\right)100$   ...... (1)

The value for $A$ is $0.917$.

Substitute $0.917$ for $A$ in equation (1).

$\begin{array}{c}T\%=\left({10}^{-A}\right)100\\ T\%=\left({10}^{\left(-0.917\right)}\right)100\\ =\left(0.121\right)100\\ =12.1\%\end{array}$

The percent transmittance for the absorbance of $0.917$ is $12.1\%$.

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=-\log T$

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9\%$.

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=-\log T$

Where,

• $A$ is the absorbance of the solution.
• $T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

${10}^{-A}={10}^{\left(\log \left(T\%/100\right)\right)}$

Simplify the expression for percent transmittance.

$T\%=\left({10}^{-A}\right)100$   ...... (1)

The value for $A$ is $0.245$.

Substitute $0.245$ for $A$ in equation (1).

$\begin{array}{c}T\%=\left({10}^{\left(-0.245\right)}\right)100\\ =\left(0.569\right)100\\ =56.9\%\end{array}$

The percent transmittance for the absorbance of $0.245$ is $56.9\%$.

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=-\log T$

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8\%$.

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=-\log T$

Where,

• $A$ is the absorbance of the solution.
• $T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

${10}^{-A}={10}^{\left(\log \left(T\%/100\right)\right)}$

Simplify the expression for percent transmittance.

$T\%=\left({10}^{-A}\right)100$   ......(1)

The value for $A$ is $0.379$.

Substitute $0.379$ for $A$ in equation (1).

$\begin{array}{c}T\%=\left({10}^{\left(-0.379\right)}\right)100\\ =\left(0.418\right)100\\ =41.8\%\end{array}$

The percent transmittance for the absorbance of $0.379$ is $41.8\%$.

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=-\log T$

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7\%$.

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=-\log T$

Where,

• $A$ is the absorbance of the solution.
• $T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

${10}^{-A}={10}^{\left(\log \left(T\%/100\right)\right)}$

Simplify the expression for percent transmittance.

$T\%=\left({10}^{-A}\right)100$   ......(1)

The value for $A$ is $0.379$.

Substitute $0.435$ for $A$ in equation (1).

$\begin{array}{c}T\%=\left({10}^{\left(-0.435\right)}\right)100\\ =\left(0.367\right)100\\ =36.7\%\end{array}$

The percent transmittance for the absorbance of $0.435$ is $36.7\%$.

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=-\log T$

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7\%$.

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=-\log T$

Where,

• $A$ is the absorbance of the solution.
• $T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

${10}^{-A}={10}^{\left(\log \left(T\%/100\right)\right)}$

Simplify the expression for percent transmittance.

$T\%=\left({10}^{-A}\right)100$   ......(1)

The value for $A$ is $0.513$.

Substitute $0.513$ for $A$ in equation (1).

$\begin{array}{c}T\%=\left({10}^{\left(-0.513\right)}\right)100\\ =\left(0.307\right)100\\ =30.7\%\end{array}$

The percent transmittance for the absorbance of $0.513$ is $30.7\%$.