Chapter 5, Problem 5.1P

To determine

The number of Hg atoms/cm³.

Explanation of Solution

Given:

Density, $\rho =13.6{\text{g/cm}}^{\text{3}}$

Atomic mass, $m=200.6\text{g/mol}$

Calculation:

The mole density of Hg is,

  $\begin{array}{l}V=\frac{\rho }{m}\\ V=\frac{13.6{\text{g/cm}}^3}{200.6\text{g/mol}}\\ V=0.067{\text{mol/cm}}^3\end{array}$

According to Avogadro’s principal, we get,

  $\begin{array}{l}1\text{mol=6}\text{.023}\times {\text{10}}^{23}\text{atoms/mol}\\ \text{0}\text{.068 mol=0}{\text{.068mol/cm}}^3\times \text{6}\text{.023}\times {\text{10}}^{23}\text{atoms/mol}\\ \text{0}\text{.068 mol=}4.095\times {10}^{22}{\text{atom/cm}}^3\end{array}$

Conclusion:

Number of Hg atoms are, $4.095\times {10}^{22}{\text{atom/cm}}^3$