Chapter 1, Problem 70QP

(a) Carbon monoxide $\left(\text{CO}\right)$ is a poisonous gas because it binds very strongly to the oxygen carrier hemoglobin in blood. A concentration of $08.0 \times {10}^2$ ppm by volume of carbon monoxide is considered lethal to humans. Calculate the volume in liters occupied by carbon monoxide in a room that measures $\text{17}\text{.6 m long}\text{. 8}\text{.80 m wide, and 2}\text{.64 m}$ high at this concentration. (b) Prolonged exposure to mercury $\left(\text{Hg}\right)$ vapor can cause neurological disorders and respiratory problems. For safe air quality control, the concentration of mercury vapor must be under $\text{0}{\text{.050 mg/m}}^{\text{3}}$. Convert this number to $\text{g/L}$. (c) The general test for type II diabetes is that the blood sugar (glucose) level should be below 120 mg per deciliter $\left(\text{mg/dt}\right)$. Convert this number to micrograms per milliliter $\text{(}\mu \text{g/mL)}$.

Interpretation Introduction

Interpretation:

The unit conversions are to be carried out by dimensional analysis for given measurements.

Concept introduction:

Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors.

Dimensional analysis is a way to convert units of measurement. In order to convert one unit to another, the relationship between those units is to be predicted. These relationships are called conversion factors.

Volume =$\text{length}\times \text{breadth}\times \text{height}$.

${\text{1 m}}^3\text{ = 1000 L}$, to convert cubic meter to liters, the conversion factor is $\frac{\text{1000 L}}{{\text{1 m}}^3}$.

${\text{1 mg = 1×10}}^{-\text{3}}\text{ g}$, to convert milligram to gram, the conversion factor is $\frac{{\text{1×10}}^{-\text{3}}\text{ g}}{\text{1 mg}}$.

$\text{1L}=\text{1000 mL}$, to convert liters to milliliters, the conversion factor is $\frac{\text{1000 mL}}{\text{1 L}}$.

$\text{1L = 10dL}$, to convert dL to L, the conversion factor is $\frac{\text{10 dL}}{\text{1 L}}$.

Answer to Problem 70QP

Solution:

a)

$\text{327 L}$

b)

${\text{5×10}}^{-8}\text{ g/L}$

c)

$1.20\times {10}^3\text{ μg/mL}$

Explanation of Solution

a) Concentration of CO $=8.00\times {10}^2\text{ ppm by volume}$.

Dimensions of the room $=17.6\text{ m }\times 8.80\text{ m }\times 2.64\text{ m}$.

Calculate the volume of the room.

The volume of the room is calculated as follows:

$\begin{array}{c}\text{Volume of the room }=\left(\text{17}\text{.6 m × 8}\text{.80 m × 2}\text{.64 m}\right)\\ =\text{408}{\text{.8832 m}}^{\text{3}}\\ =\text{4}{\text{.089×10}}^2{\text{ m}}^3\end{array}$

Now, calculate the volume occupied by CO in liters.

Set up the unit conversion calculation using the appropriate conversion factor.

The conversion is as follows:

$\left(\text{4}\text{.089}\times {\text{10}}^2\cancel{{\text{m}}^3}\right)\left(\frac{8.00\times {10}^2}{{\text{10}}^6}\right)\left(\frac{{\text{10}}^3\text{ L}}{1\cancel{{\text{m}}^3}}\right)=\text{ 327 L}$

The units in the conversion factor must cancel to give the correct unit for the answer.

Hence, the volume occupied by CO in the room is $\text{327 L}$.

b) The concentration of mercury vapor must be under $\text{0}{\text{.050 mg/m}}^{\text{3}}$

Given information:.

Convert the quantity $\text{0}{\text{.050 mg/m}}^{\text{3}}$ to g/L.

Set up the unit conversion calculation using the appropriate conversion factors.

The conversion is as follows:

$\left(\text{0}\text{.050 }\cancel{\text{mg}}\text{/}\cancel{{\text{m}}^{\text{3}}}\right)\left(\frac{\text{1 g}}{\text{1000 }\cancel{\text{mg}}}\right)\left(\frac{\text{1 }\cancel{{\text{m}}^3}}{{\text{10}}^3\text{ L}}\right)={\text{5×10}}^{-8}\text{ g/L}$

The units in the conversion factor must cancel to give the correct unit for the answer.

Hence,

$\text{0}{\text{.050 mg/m}}^{\text{3}}={\text{5×10}}^{-8}\text{ g/L}$

c) Blood sugar level $=120\text{ mg/dL}$

Convert the quantity $\text{120 mg/dL}$ to $\text{μg/mL}$.

Set up the unit conversion calculation using the appropriate conversion factors.

The conversion is as follows:

$\left(\text{120 }\cancel{\text{mg}}\text{/}\cancel{\text{dL}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{ μg}}{\text{1 }\cancel{\text{mg}}}\right)\left(\frac{\text{10 }\cancel{\text{dL}}}{\text{1 }\cancel{\text{L}}}\right)\left(\frac{\text{1}\cancel{\text{ L}}}{\text{1000 mL}}\right)\text{ = 1}\text{.20}\times {\text{10}}^3\text{ μg/mL}$

The units in the conversion factor must cancel to give the correct unit for the answer.

Hence,

$\text{120 mg/dL}=\text{1}\text{.20}\times {\text{10}}^3\text{ μg/mL}$.