Question 1.34 - Chemistry: Molecular Nature Of Matter And Change

Question

The average density of Earth is 5.52 g/cm³. What is its density in (a) km/m³ ; (b) lb/ft³ ?

Step-by-Step Solution

Answer

Answer

 

(a) The average density of Earth is $5.52\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$ 

(b) The average density of Earth is $345 \mathrm{lb}/{\mathrm{ft}}^3$ .

Convert the density from g/cm 3 to kg/m 3 .

$$\begin{gathered} \mathrm{The} \mathrm{unit} \mathrm{conversion} \mathrm{equation} \mathrm{that} \mathrm{relates} \mathrm{from} \mathrm{the} \mathrm{grams} \mathrm{to} \mathrm{kilograms} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} \mathrm{follows}: \\ 1 \mathrm{kg} = {10}^3\mathrm{g} \\ \mathrm{The} \mathrm{unit} \mathrm{conversion} \mathrm{factor} \mathrm{for} \mathrm{converting} \mathrm{the} \mathrm{mass} \mathrm{from} \mathrm{grams} \mathrm{to} \mathrm{kilograms} \mathrm{is} \\ \frac{1 \mathrm{kg}}{{10}^3\mathrm{g}} \\ \mathrm{The} \mathrm{unit} \mathrm{conversion} \mathrm{equation} \mathrm{that} \mathrm{describes} \mathrm{the} \mathrm{relation} \mathrm{between} \mathrm{centimeters} \mathrm{and} \mathrm{meters} \mathrm{can} \mathrm{be} \mathrm{shown} \mathrm{as} \mathrm{follows}: \\ 1\mathrm{m} = {10}^2\mathrm{cm} \\ \mathrm{The} \mathrm{conversion} \mathrm{factor} \mathrm{needed} \mathrm{to} \mathrm{convert} \mathrm{meters} \mathrm{to} \mathrm{centimeters} \mathrm{is} \mathrm{as} \mathrm{follows}: \\ \frac{{10}^2\mathrm{cm}}{1\mathrm{m}} \\ \mathrm{Now}, \mathrm{use} \mathrm{the} \mathrm{above} \mathrm{two} \mathrm{conversion} \mathrm{factors} \mathrm{in} \mathrm{any} \mathrm{order} \mathrm{to} \mathrm{convert} \mathrm{the} \mathrm{density} \mathrm{of} \mathrm{Earth} \mathrm{from} \mathrm{g}/\mathrm{cm}3 \mathrm{to} \mathrm{kg}/\mathrm{m}3 \mathrm{as} \mathrm{shown} \mathrm{below}: \\ \frac{5.52g}{1c{m}^3} \times \frac{1 kg}{{10}^{3}g} \times {\left(\frac{{10}^{2}cm}{1m}\right)}^3 = 5.52\times {10}^3 kg/m^3 \\ \mathrm{Hence}, \mathrm{the} \mathrm{average} \mathrm{density} \mathrm{of} \mathrm{Earth} \mathrm{in} \mathrm{the} \mathrm{SI} \mathrm{units} \mathrm{of} \mathrm{is} 5.52\times {10}^3 \mathrm{kg}/{\mathrm{m}}^3 . \end{gathered}$$

Convert the density from g/cm 3 to lb/ft 3 .

The unit conversion equation that relates the units of mass between pounds and kilograms can be written as follows:

1kg = 2.205 lb

The conversion factor required to convert kilograms to pounds is as follows:

$\frac{2.205lb}{1kg}$

The unit conversion equation for converting centimeters to inches can be shown as follows:

 1 inch = 2.54cm

The conversion ratio of inches to centimeters is as follows:

$\frac{2.54cm}{1inch}$

By using this unit conversion equation 1ft = 12in , the unit conversion factor for converting feet to inches is shown below:

$\frac{12in}{1ft}$

The unit analysis setup for this conversion can be written as follows:

$\frac{5.52g}{1c{m}^3} \times \frac{1kg}{{10}^{3}g} \times \frac{2.205lb}{1kg} \times {\left(\frac{2.54cm}{1in}\right)}^3 \times {\left(\frac{12in}{1ft}\right)}^3 = 345 lb/f{t}^3$

Hence, the average density of Earth in the non-SI units of  is  345 lb/ft³.