Q48P - Fundamentals Of Physics

Question

The common Eastern mole, a mammal, typically has a mass of ${}^{\mathbf{75}\mathbf{ }\mathbf{g}}$ , which corresponds to about ${}^{\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{moles}}$ of atoms. (A mole of atoms is${}^{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{ }\mathbf{atoms}}$.) In atomic mass units (u), what is the average mass of the atoms in the common Eastern mole?

Step-by-Step Solution

Answer

The average mass of the atoms in the common Eastern mole is${}^{\mathbf{10}\mathbf{ }\mathbf{u}}$ .

Step 1: Given data

A mole of atoms corresponds to ${}^{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{ }\mathbf{atoms}}$ .

${}^{\mathbf{75}\mathbf{ }\mathbf{g}}$ of common Eastern mole corresponds to ${}^{\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{atoms}}$ .

Step 2: Understanding the conversion factor

Use the given conversion factor to calculate the number of moles in a common easter mole. Using this number and the number of atoms in mole, calculate the average mass of one atom in the common eastern mole.

Step 3: Determination of the mass of one mole of atom in a common eastern mole

The mass of one mole of atoms in the common Eastern mole ${}^{{\mathbf{m}}_{\mathbf{c}}}$ is,

 $$\begin{gathered} {\mathit{m}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{75}}{\mathbf{7}\mathbf{.}\mathbf{5}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{10}\mathbf{ }\mathit{g} \end{gathered}$$

Thus, the mass of one mole of atom in the common eastern mole is ${}^{\mathbf{10}\mathbf{ }\mathbf{g}}$ .

Step 4: Determination of the mass of one atom in a common eastern mole

The average mass of one atom in the common Eastern mole is ${}^{{\mathbf{m}}_{\mathbf{c}}}$ divided by Avogadro’s Number $N_A$.

$$\begin{gathered} \frac{{\mathbf{m}}_{\mathbf{c}}}{{\mathbf{N}}_{\mathbf{A}}}\mathbf{=}\frac{\mathbf{10}\mathbf{ }\mathbf{g}}{\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{66}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{10}\mathbf{ }\mathit{u} \end{gathered}$$

Thus, the average mass of one atom in the common eastern mode is ${}^{\mathbf{10}\mathbf{ }\mathbf{u}}$ .