Chapter 1, Problem 1.99QP

Interpretation Introduction

Interpretation: From the given data, density of the pheromone should be determined.

Concept Introduction:

• $\begin{array}{l}Volumeofacylinder=area\times height\\ =\pi r^{2}h\end{array}$
• $Density=\frac{mass}{Volume}$
• Conversion formula:

  $\begin{array}{l}\text{1}\text{mile}\text{=}\text{1609}\text{metre}\\ \text{1}\text{centimetre}\text{=}\text{0}\text{.01}\text{metre}\end{array}$

  $\begin{array}{l}\text{1}\text{ft}\text{=}\text{12}\text{in}\\ \text{1}\text{in}\text{=}\text{2}\text{.54}\text{cm}\end{array}$

  $\begin{array}{l}\text{1}{\text{cm}}^{\text{3}}\text{=1mL}\\ \text{1}\text{L}\text{=}\text{1000}\text{m}\text{L}\end{array}$

Answer to Problem 1.99QP

Answer

Density $=4.0\times 10^{-19}g/L$

Explanation of Solution

Explanation

To determine: density of the pheromone s

To determine the density of the pheromone s, following steps should be followed,

To convert: The given radius and height from mile to centimeter and from foot to centimeter respectively.

$\begin{array}{l}\text{Radius}\text{=8}\text{.05}{\text{×10}}^{\text{4}}\text{cm}\\ \text{Height=}\text{1}\text{.22}\text{×}{\text{10}}^{\text{3}}\text{cm}\end{array}$

The given radius is $\text{0}\text{.50}\text{mi}$. The given radius can be converted in to centimeter by using the given conversion formula, that is,

$\begin{array}{l}\text{1}\text{mile}\text{=}\text{1609}\text{metre}\\ \text{1}\text{centimetre}\text{=}\text{0}\text{.01}\text{metre}\end{array}$

$\text{0}\text{.50}\text{mi}\text{×}\frac{\text{1609}\text{m}}{\text{1}\text{mi}}\text{×}\frac{\text{1}\text{cm}}{\text{0}\text{.01}\text{m}}\text{=}\text{8}\text{.05}{\text{×10}}^{\text{4}}\text{cm}$

The given height in foot is $\text{40}\text{ft}$, the given height can be converted to centimeter by using the given formula;

$\begin{array}{l}\text{1}\text{ft}\text{=}\text{12}\text{in}\\ \text{1}\text{in}\text{=}\text{2}\text{.54}\text{cm}\end{array}$

$\text{40}\text{ft}\text{× }\frac{\text{12}\text{in}}{\text{1}\text{ft}}\text{×}\frac{\text{2}\text{.54}\text{cm}}{\text{1in}}\text{=}\text{1}\text{.22}\text{×}{\text{10}}^{\text{3}}\text{cm}$

To determine the volume of a cylinder

$\text{Volume}\text{=}\text{2}\text{.48}{\text{×10}}^{\text{10}}\text{L}$

Volume of cylinder can be determined by using the given formula. That is,

$\begin{array}{l}Volumeofacylinder=area\times height\\ =\pi r^{2}h\end{array}$

$\begin{array}{l}\text{Here,}\\ \text{Radius}\text{(r)}\text{=}\text{8}\text{.05}\text{×}{\text{10}}^{\text{4}}\text{cm}\\ \text{Height}\text{(h)}\text{=}\text{1}\text{.22}\text{×}{\text{10}}^{\text{3}}\text{cm}\\ \text{Volume}\text{of}\text{a}\text{cylinder}\text{=}\text{π}{\text{r}}^{\text{2}}\text{h}\\ \text{π}\text{=3}\text{.14}\\ \text{Volume}\text{of}\text{a}\text{cylinder}\text{=3}\text{.14×(8}\text{.05}\text{×}{\text{10}}^{\text{4}}{\text{cm)}}^{\text{2}}\text{×}\text{(1}\text{.22}\text{×}{\text{10}}^{\text{3}}\text{cm)}\\ \text{=}\text{2}\text{.48}{\text{10}}^{\text{13}}{\text{cm}}^{\text{3}}\end{array}$

It is needed to convert the volume to litters by using the given conversion formula.

$\begin{array}{l}\text{1}{\text{cm}}^{\text{3}}\text{=1mL}\\ \text{1}\text{L}\text{=}\text{1000}\text{m}\text{L}\end{array}$

$\begin{array}{l}\text{Here,}\\ \text{Radius}\text{(r)}\text{=}\text{8}\text{.05}\text{×}{\text{10}}^{\text{4}}\text{cm}\\ \text{Height}\text{(h)}\text{=}\text{1}\text{.22}\text{×}{\text{10}}^{\text{3}}\text{cm}\\ \text{Volume}\text{of}\text{a}\text{cylinder}\text{=}\text{π}{\text{r}}^{\text{2}}\text{h}\\ \text{π}\text{=3}\text{.14}\\ \text{Volume}\text{of}\text{a}\text{cylinder}\text{=3}\text{.14×(8}\text{.05}\text{×}{\text{10}}^{\text{4}}{\text{cm)}}^{\text{2}}\text{×}\text{(1}\text{.22}\text{×}{\text{10}}^{\text{3}}\text{cm)}\\ \text{=}\text{2}\text{.48}{\text{10}}^{\text{13}}{\text{cm}}^{\text{3}}\\ \text{2}\text{.48}{\text{10}}^{\text{13}}{\text{cm}}^{\text{3}}\text{×}\frac{\text{1}\text{mL}}{\text{1}{\text{cm}}^{\text{3}}}\text{×}\frac{\text{1L}}{\text{1000}\text{mL}}\text{=}\text{2}\text{.48}\text{×}{\text{10}}^{\text{10}}\text{L}\end{array}$

To determine: Density of the pheromone

$\begin{array}{l}\text{4}\text{.0}\text{×}{\text{10}}^{\text{-19}}\text{g/L}\\ \end{array}$

Density of the pheromone can be determined by using the given formula,

$Density=\frac{mass}{Volume}$

Here,

$\begin{array}{l}\text{Mass=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-8}}\text{g}\\ \text{Volume=}\text{2}\text{.48}\text{×}{\text{10}}^{\text{10}}\text{L}\\ \text{Density}\text{=}\frac{\text{1}\text{.0}\text{×}{\text{10}}^{\text{-8}}\text{g}}{\text{2}\text{.48}\text{×}{\text{10}}^{\text{10}}\text{L}}\text{=}\text{4}\text{.0}\text{×}{\text{10}}^{\text{-19}}\text{g/L}\end{array}$

Conclusion

Conclusion

From the given data, density of the pheromone is determined.