Chapter 17, Problem 45R

To determine

To Find: The density of a cube of lead.

Answer to Problem 45R

  $11.34{\text{g/cm}}^{\text{3}}$

Explanation of Solution

Given Information:

Mass of cube = $11.34\text{kg}\text{=}1134\text{0}\text{g}$

Side of cube = $10\text{cm}$

Volume = ${(10)}^3\text{cm}$

Formula used:

  $\rho =\frac{m}{V}$

Where,

  M = mass of the cube

  V = volume the cube

  $\rho$ = density

Calculation:

Substitute the values and solve:

  $\begin{array}{l}\rho =\frac{11340\text{g}}{{\text{(10}\text{cm)}}^{\text{3}}}\\ \rho =\frac{11340\text{g}}{1000{\text{cm}}^3}\\ \rho =11.34{\text{g/cm}}^3\end{array}$

Conclusion: The density is $11.34{\text{g/cm}}^{\text{3}}$.