Chapter 1, Problem 1.98QP

(a)

Interpretation Introduction

Interpretation:

The volume in liters occupied by carbon monoxide in a room that measures $17.6long,8.80mwideand2.64mhigh$ has to be calculated.

Concept Introduction:

Volume can be calculated by the equation given below:

$\text{length×}\text{width×}\text{height}$

Volume is the ratio of mass to density and its SI unit is ${\text{m}}^{\text{3}}$.

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

Unit conversions:

$\begin{array}{l}1L=1000c{m}^3\\ \\ 1m=100cm\end{array}$

Explanation of Solution

Given data:

$\begin{array}{c}concentration=8.00\times {10}^{2}ppm\\ length=17.6m\\ width=8.80m\\ height=2.64m\end{array}$

Volume can be calculated by the equation given below:

$\text{length×}\text{width×}\text{height}$

$\begin{array}{c}\text{V}\text{=}\text{length×}\text{width×}\text{height}\\ \text{=}\text{17}\text{.6}\text{m×}\text{8}\text{.80}\text{m×}\text{2}\text{.64}\text{m}\\ \text{=}\text{409}{\text{m}}^{\text{3}}\end{array}$

Volume has to be converted into liters

$\text{409}{\text{m}}^{\text{3}}\text{×}{\left(\frac{\text{1}\text{cm}}{{\text{1×10}}^{\text{-2}}\text{m}}\right)}^{\text{3}}\text{×}\frac{\text{1}\text{L}}{\text{1000}{\text{cm}}^{\text{3}}}\text{=}\text{4}{\text{.09×10}}^{\text{5}}\text{L}\text{air}$

The concentration is given by $8.00\times {10}^{2}ppm$

The volume in liters can be calculated as follows:

$4.09\times {10}^{5}Lair\times \frac{8.00\times {10}^{2}LCO}{1\times {10}^{6}Lair}=327LCO$

(b)

Interpretation Introduction

Interpretation:

The given concentration has to be converted from $mg/m^{3}tog/L$

Concept Introduction

Unit conversions:

$\begin{array}{l}1g=1000mg\\ \\ 1m=100cm\\ \\ 1L=1000c{m}^3\end{array}$

Explanation of Solution

Given data:

$concentration=0.050mg/m^3$

The concentration has to be converted from $mg/m^{3}tog/L$.

The unit conversions useful for this are given below:

$\begin{array}{l}1g=1000mg\\ \\ 1m=100cm\\ \\ 1L=1000c{m}^3\end{array}$

$\frac{0.050mg}{1m^3}\times \frac{1\times {10}^{-3}g}{1mg}\times {\left(\frac{1\times {10}^{-2}m}{1cm}\right)}^3\times \frac{1000c{m}^3}{1L}=5.0\times {10}^{-8}g/L$

(c)

Interpretation Introduction

Interpretation:

The given blood sugar level has to be converted from $mg/dLto\mu g/mL$

Concept Introduction

Unit conversions:

$\begin{array}{l}1\mu g=1\times {10}^{-3}g\\ \\ 1mL=1\times {10}^{-2}dL\end{array}$

Explanation of Solution

Given data:

$concentration=120mg/dL$

The concentration has to be converted from $mg/dLto\mu g/mL$.

The unit conversions useful for this are given below:

$\begin{array}{l}1\mu g=1\times {10}^{-3}g\\ \\ 1mL=1\times {10}^{-2}dL\end{array}$

$\frac{120mg}{1dL}\times \frac{1\mu g}{1\times {10}^{-3}mg}\times \left(\frac{1\times {10}^{-2}dL}{1dL}\right)=1.20\times {10}^3\mu g/L$