Q37P - Fundamentals Of Physics

Question

A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (${}^{\mathbf{One}\mathbf{ }\mathbf{mole}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}\mathbf{ }\mathbf{units}}}$ .)

Step-by-Step Solution

Answer

The edge length of cubical box is ${}^{\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathbf{m}}$. 

Step 1: Given data

The edge length of the sugar cube is${}^{\mathbf{1}\mathbf{ }\mathbf{cm}}$.

$\mathbf{1}\mathbf{ }\mathbf{mole}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}\mathbf{ }}\mathbf{ }\mathbf{units}$ 

Step 2: Understanding the unit conversion

In this problem, the volume of one mole of sugar cubes will be equal to the cubical box that contains one mole of sugar cubes. 

The volume of a cube is given as: 

$\mathrm{V}={\mathrm{a}}^3$                                                            … (i)

Here, ${}^a$  is the edge length of cube.

Step 3: Determination of the volume of one mole of sugar cube

Using equation (i), the volume of one sugar cube is calculated as: 

$$\begin{gathered} {\mathbf{V}}_{\mathbf{sc}}\mathbf{ }\mathbf{=}{\left(1.0 \mathrm{cm}\right)}^{\mathbf{3}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}{\left(1.0 \mathrm{cm} \times \frac{{10}^{-2} m}{1 \mathrm{cm}}\right)}^{\mathbf{3}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }{\mathbf{m}}^{\mathbf{3}} \end{gathered}$$

As we know,

$\mathbf{1}\mathbf{ }\mathbf{mole}\mathbf{ }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}\mathbf{ }}\mathbf{units}$.

The volume of the cubical box is equal to the volume of one mole of sugar cubes. Therefore, volume of cubical box is, 

$$\begin{gathered} \mathbf{V}\mathbf{ }\mathbf{=}\mathbf{ }\left(1.0\times {10}^{-6} m^3\right)\mathbf{\times }\left(6.02\times {10}^{23}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}\mathbf{ }}{\mathbf{m}}^{\mathbf{3}} \end{gathered}$$

Thus, the volume of cubical box is ${}^{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}\mathbf{ }}{\mathbf{m}}^{\mathbf{3}}}$ .

Step 4: Determination of the edge length of the cubical box

Using equation (i), the edge length of cubical box is calculated as: 

$$\begin{gathered} \mathbf{a}\mathbf{\text{'}}\mathbf{ }\mathbf{=}\mathbf{ }\sqrt[\mathbf{3}]{\mathbf{V}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\sqrt[\mathbf{3}]{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{17}\mathbf{ }}{\mathbf{m}}^{\mathbf{3}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{\times }\mathbf{ }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathbf{m} \end{gathered}$$

Thus, the edge length of cubical box is ${}^{\mathbf{ }\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{\times }\mathbf{ }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathbf{m}}$ .