Chapter 1, Problem 1.115QP

(a)

Interpretation Introduction

Interpretation:

The storage temperature of cidofovir should be converted from Fahrenheit to Celsius.  The density of the fluid in the single vial of the cidofovir should be determined.  The density of the fluid in the single vial of cidofovir should be converted to $\text{g/L}$ and to ${\text{kg/m}}^3$.  The dosage of the cidofovir to a $\text{185}\text{lb}$ man should be determined.

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

${\text{F}}^{\text{o}}\text{=}\left({}^{\text{o}}\text{C×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}$

• Conversional formula from Celsius to Kelvin,

  $\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

• Equation for density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

To convert: the storage temperature of cidofovir from Fahrenheit to Celsius

The given storage temperature of cidofovir is in Fahrenheit.

Explanation of Solution

Given,

The given storage temperature of cidofovir is ${\text{68}}^{\text{0}}\text{F}-{\text{77}}^{\text{0}}\text{F}$

Conversional formulafrom Fahrenheit toCelsius,

${\text{F}}^{\text{o}}\text{=}\left({}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}$

Therefore,

${}^{\text{0}}\text{C}\text{=}\left({}^{\text{0}}\text{F}-{\text{32}}^{\text{0}}\text{F}\right)\frac{{\text{5}}^{\text{0}}\text{C}}{{\text{9}}^{\text{0}}\text{F}}$

Converting ${\text{68}}^{\text{0}}\text{F}$ to Celsius,

$\begin{array}{c}{}^{\text{0}}\text{C}=\left({68}^{\text{0}}\text{F}-{\text{32}}^{\text{0}}\text{F}\right)\frac{{\text{5}}^{\text{0}}\text{C}}{{\text{9}}^{\text{0}}\text{F}}\\ ={20}^0\text{C}\end{array}$

Converting ${\text{77}}^{\text{0}}\text{F}$ to Celsius,

$\begin{array}{c}{}^{\text{0}}\text{C}=\left({\text{77}}^{\text{0}}\text{F}-{\text{32}}^{\text{0}}\text{F}\right)\frac{{\text{5}}^{\text{0}}\text{C}}{{\text{9}}^{\text{0}}\text{F}}\\ ={25}^0\text{C}\end{array}$

The storage temperature of cidofovir in Celsiusis ${20}^0\text{C}-{25}^0\text{C}$.

(b)

Interpretation Introduction

Interpretation:

The storage temperature of cidofovir should be converted from Fahrenheit to Celsius.  The density of the fluid in the single vial of the cidofovir should be determined.  The density of the fluid in the single vial of cidofovir should be converted to $\text{g/L}$ and to ${\text{kg/m}}^3$.  The dosage of the cidofovir to a $\text{185}\text{lb}$ man should be determined.

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

${\text{F}}^{\text{o}}\text{=}\left({}^{\text{o}}\text{C×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}$

• Conversional formula from Celsius to Kelvin,

  $\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

• Equation for density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

To determine: The density of the fluid in the single vial of the cidofovir

Explanation of Solution

Given,

The volume of liquid in the single vial of the cidofovir is $\text{5}\text{.0}\text{mL}$

The mass of liquid in the single vial of the cidofovir is $\text{5}\text{.89 g}$

Equationfor density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Therefore the density of the fluid in the single vial of the cidofovir is,

$\begin{array}{c}\text{Density}\text{=}\frac{\text{5}\text{.89}\text{g}}{\text{5}\text{.0}\text{mL}}\\ \text{=}\text{1}\text{.18}\text{g/mL}\end{array}$

The density of the fluid in the single vial of the cidofovir is $\text{1}\text{.18}\text{g/mL}$

(c)

Interpretation Introduction

Interpretation:

The storage temperature of cidofovir should be converted from Fahrenheit to Celsius.  The density of the fluid in the single vial of the cidofovir should be determined.  The density of the fluid in the single vial of cidofovir should be converted to $\text{g/L}$ and to ${\text{kg/m}}^3$.  The dosage of the cidofovir to a $\text{185}\text{lb}$ man should be determined.

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

${\text{F}}^{\text{o}}\text{=}\left({}^{\text{o}}\text{C×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}$

• Conversional formula from Celsius to Kelvin,

  $\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

• Equation for density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

The given density of the fluid in the single vial is in $\text{g/mL}$.

Explanation of Solution

Given,

The density of the fluid in the single vial of the cidofovir is $\text{1}\text{.18}\text{g/mL}$

To convert $\text{mL}$ to $\text{L}$,multiple with $\frac{\text{1}\text{×}{\text{10}}^{\text{3}}\text{mL}}{\text{1}\text{L}}$

Converting $\text{g/mL}$ to ${\text{kg/m}}^3$,

$\text{1}\text{g/mL}=1000{\text{kg/m}}^3$

Substituting these values

The density of the fluid in the single vial of the cidofovir in $\text{g/L}$ is,

$\frac{\text{1}\text{.18}\text{g/mL}}{\text{mL}}\text{×}\frac{\text{1}\text{×}{\text{10}}^{\text{3}}\text{mL}}{\text{1}\text{L}}\text{=}\text{1180}\text{g/L}$

The density of the fluid in the single vial of the cidofovir in ${\text{kg/m}}^3$ is,

$\text{1}\text{.18}\text{g/mL}\text{×}\frac{\text{1000}{\text{kg/m}}^{\text{3}}}{\text{1}\text{g/mL}}\text{=}\text{1180}{\text{kg/m}}^{\text{3}}$

The density of the fluid in the single vial of the cidofovir in $\text{g/L}$ is $\text{1180}\text{g/L}$

The density of the fluid in the single vial of the cidofovir in ${\text{kg/m}}^3$ is $\text{1180}{\text{kg/m}}^{\text{3}}$

(d)

Interpretation Introduction

Interpretation:

The storage temperature of cidofovir should be converted from Fahrenheit to Celsius.  The density of the fluid in the single vial of the cidofovir should be determined.  The density of the fluid in the single vial of cidofovir should be converted to $\text{g/L}$ and to ${\text{kg/m}}^3$.  The dosage of the cidofovir to a $\text{185}\text{lb}$ man should be determined.

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

${\text{F}}^{\text{o}}\text{=}\left({}^{\text{o}}\text{C×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}$

• Conversional formula from Celsius to Kelvin,

  $\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

• Equation for density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

To determine: The dosage of the cidofovir to a $\text{185}\text{lb}$ man

Explanation of Solution

Given,

Body weight of man is $\text{185}\text{lb}$

Here $\text{lb}$ is first converted to $\text{g}$ and then to $\text{kg}$,

Weight of the man is,

$\text{185}\text{lb}\text{×}\frac{\text{453}\text{.6}\text{g}}{\text{1}\text{lb}}\text{×}\frac{\text{1}\text{kg}}{\text{1}\text{×}{\text{10}}^{\text{3}}\text{g}}\text{=}\text{83}\text{.92}\text{kg}$

Therefore, the dosage of the cidofovir to a $\text{185}\text{lb}$ man is,

$\frac{\text{5}\text{mg}}{\text{kg}}\text{×}\text{83}\text{.92}\text{kg}\text{=}\text{420}\text{mg}$

The dosage of the cidofovir to a $\text{185}\text{lb}$ man is $\text{420}\text{mg}$