Chapter 3.3, Problem 48SP

Interpretation Introduction

Interpretation : The density in kilograms per cubic meter is to be expressed.

Concept Introduction : An equivalent measurement ratio is referred to as a conversion factor. An effective approach for resolving conversion problems where measurement in one unit is changed to an equivalent measurement in another is dimensional analysis.

Answer to Problem 48SP

The density in kilograms per cubic meter is $1.93\times {10}^4{\text{kg/m}}^3$ .

Explanation of Solution

Given information :

The density of gold is $19.3{\text{g/cm}}^3$ .

It is known that $\text{1}\text{kg}\text{=}{\text{10}}^3\text{g}$ and $\text{1}{\text{m}}^3\text{=}{\text{10}}^6{\text{cm}}^3$ .

An equivalent measurement ratio is referred to as a conversion factor. An effective approach for resolving conversion problems where measurement in one unit is changed to an equivalent measurement in another is dimensional analysis.

The desired conversion is $\text{g}\to \text{kg}$ and ${\text{cm}}^3\to {\text{m}}^3$ .

The conversion of the density is given as:

  $\begin{array}{c}19.3{\text{g/cm}}^3\text{=}\frac{19.3\cancel{\text{g}}}{\cancel{{\text{cm}}^3}}\times \left(\frac{{10}^{-3}\text{kg}}{1\cancel{\text{g}}}\right)\times \left(\frac{1\cancel{{\text{cm}}^3}}{{10}^{-6}{\text{m}}^3}\right)\left(\because \text{1}\text{kg}\text{=}{\text{10}}^3\text{g}\text{,}\text{1}{\text{m}}^3\text{=}{\text{10}}^6{\text{cm}}^3\right)\\ =19.3\times {10}^3{\text{kg/m}}^3\\ =1.93\times {10}^4{\text{kg/m}}^3\end{array}$

Conclusion

The density in kilograms per cubic meter is $1.93\times {10}^4{\text{kg/m}}^3$ .