Chapter 1, Problem 55QP

Carry out the following conversions: (a) $\text{242 lb}$ to milligrams, (b) $\text{68}{\text{.3 cm}}^{\text{3}}$ to cubic meters, (c) $\text{7}{\text{.2 m}}^{\text{3}}$ to liters, (d) 28.3 $\text{μg}$ to pounds.

Interpretation Introduction

Interpretation:

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

Concept introduction:

Dimensional analysis is a way to convert the units of measurement. In order to convert one unit to another, one needs to know the relationship between those units. These relationships are called conversion factors. Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors.

Conversion factor is a fraction obtained from a relationship between the units. It is written as a ratio, and can be inverted to give two conversion factors for every relationship.

$\text{1 lb = 453}\text{.6 g}$, to convert lb to grams, the conversion factor is $\frac{\text{453}\text{.6 g}}{\text{1 lb}}$.

$\text{1 g = 1000 mg}$, to convert g to milligram, the conversion factor is $\frac{\text{1000 mg}}{\text{1 g}}$.

$\text{1 m = 100 cm}$, to convert centimeter to meter, the conversion factor is $\frac{\text{1 m}}{\text{100 cm}}$.

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

$\text{1 μg = 1}\times {\text{10}}^{-6}\text{ g}$, to convert $\text{μg}$ to gram, the conversion factor is $\frac{1\times {10}^{-3}\text{g}}{\text{1 μg}}$.

$\text{1 lb = 453}\text{.6 g}$, to convert grams to lb, the conversion factor is $\frac{\text{1 lb}}{\text{453}\text{.6 g}}$.

Answer to Problem 55QP

Solution:

a)

$1.10\times {10}^5\text{ mg}$

b)

$\text{6}{\text{.83×10}}^{\text{-5}}{\text{ m}}^{\text{3}}$

c)

$\text{7}\text{.2}\times {\text{10}}^3\text{ L}$

d)

$\text{6}{\text{.24×10}}^{-5}\text{ lb}$

Explanation of Solution

a) $\text{242 lb}$ to milligram.

Convert the quantity $\text{242 lb}$ to milligrams.

The conversion factors for the relationship $\text{1 lb = 453}\text{.6 g}$ are as follows:

$\frac{\text{1 lb}}{\text{453}\text{.6 g}}$ or $\frac{\text{453}\text{.6 g}}{\text{1 lb}}$

The conversion factors for the relationship $\text{1 g = 1000 mg}$ are as follows:

$\frac{\text{1 g}}{1000\text{ mg}}$ or $\frac{\text{1000 mg}}{\text{1 g}}$

Now, set up the unit conversion problem using the appropriate conversion factors.

$\left(\text{242 }\cancel{\text{lb}}\text{ × }\frac{\text{453}\text{.6 }\cancel{\text{g}}}{\text{1 }\cancel{\text{lb}}}\text{ × }\frac{\text{1000 mg}}{\text{1 }\cancel{\text{g}}}\right)=\text{1}{\text{.09×10}}^5\text{ mg}$

The units in the conversion factor must cancel to give the correct unit for the answer. The answer is rounded off to $1.10\times {10}^5\text{ mg}$.

Hence,

$\text{242lb = }1.10\times {10}^5\text{ mg}$.

b) $\text{68}{\text{.3 cm}}^{\text{3}}$ to cubic meters.

Convert the quantity $\text{68}{\text{.3 cm}}^{\text{3}}$ to cubic meters.

The conversion factors for the relationship $\text{1 m = 100 cm}$ are as follows:

$\frac{\text{1 m}}{\text{100 cm}}$ or $\frac{\text{100 cm}}{\text{1 m}}$

Now, set up the unit conversion problem using the appropriate conversion factor.

$\left(\text{68}\text{.3 }\cancel{{\text{cm}}^3}\text{ × }{\left(\frac{\text{1 m}}{\text{100 }\cancel{\text{cm}}}\right)}^3\right)=\text{6}\text{.83}\times {\text{10}}^{-5}{\text{ m}}^3$

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

Hence,

$\text{68}{\text{.3 cm}}^{\text{3}}=\text{6}{\text{.83×10}}^{-\text{5}}{\text{ m}}^{\text{3}}$

c) $\text{7}{\text{.2 m}}^{\text{3}}$ to liters.

Convert the quantity $\text{7}{\text{.2 m}}^{\text{3}}$ to liters.

The conversion factors for the relationship ${\text{1 m}}^3\text{ = 1000 L}$ are as follows:

$\frac{{\text{1 m}}^3}{\text{1000 L}}$ or $\frac{\text{1000 L}}{{\text{1 m}}^3}$

Now, set up the unit conversion problem using the appropriate conversion factor.

$\left(\text{7}\text{.2 }\cancel{{\text{m}}^3}\text{ × }\frac{\text{1000 L}}{\text{1 }\cancel{{\text{m}}^3}}\right)=\text{7200 L}$

The units in the conversion factor must cancel to give the correct unit for the answer. The answer is rounded off to $\text{7}\text{.2}\times {\text{10}}^3\text{ L}$.

Hence,

$\text{7}{\text{.2 m}}^3\text{ = 7}\text{.2}\times {\text{10}}^3\text{ L}$.

d) $\text{28}\text{.3 μg}$ to pounds (lb).

Convert the quantity $\text{28}\text{.3 μg}$ to pounds (lb).

The conversion factors for the relationship $\text{1 μg = 1}\times {\text{10}}^{-6}\text{ g}$ are as follows:

$\frac{\text{1 μg}}{1\times {10}^{-3}\text{g}}$ or $\frac{1\times {10}^{-3}\text{g}}{\text{1 μg}}$

The conversion factors for the relationship $\text{453}\text{.6 g = 1 lb}$ are as follows:

$\frac{\text{453}\text{.6 g}}{\text{1 lb}}$ or $\frac{\text{1 lb}}{\text{453}\text{.6 g}}$

Now, set up the unit conversion problem using the appropriate conversion factors.

$\left(\text{28}\text{.3 }\cancel{\text{μg}}\text{ × }\frac{1\times {10}^{-3}\cancel{\text{g}}}{\text{1 }\cancel{\text{μg}}}\text{ × }\frac{\text{1 lb}}{\text{453}\text{.6 }\cancel{\text{g}}}\right)=\text{6}{\text{.24×10}}^{-5}\text{ lb}$

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

Hence,

$\text{28}\text{.3 μg = 6}{\text{.24×10}}^{-5}\text{ lb}$.