Chapter 11, Problem 11.148QP

Interpretation Introduction

Interpretation:

From the calculation of radius and density the unknown alkali metal has to be identified.

Concept Introduction:

The radius of a body-centered cubic unit cell:

$\begin{array}{l}\text{r}\text{=}\frac{\sqrt{\text{3}}}{\text{4}}\times \text{a}\\ \text{a}\text{=}\text{edge}\text{length}\end{array}$

Volume of the cubic unit cell:

$\text{V}\text{=}{\text{a}}^{\text{3}}$

The edge length of the cubic unit cell:

$\text{a}=\sqrt[3]{\text{V}}$

Answer to Problem 11.148QP

The unknown alkali-metal is found to be Sodium.

Explanation of Solution

Given data:

Vapor pressure $\left(\text{P}\right)$ : $\text{19}\text{.2}\text{mmHg}$

Temperature $\left(\text{T}\right)$:$\text{1235K}$

Volume $\left(\text{V}\right)$: $\text{0}\text{.843L}$

It is known that

$\text{R}\text{=}\text{Gas}\text{constant}\text{=}\text{0}\text{.0821}\text{L}\text{.}\text{atm}$

Calculating the number of moles of the unknown alkali metal:

The ideal gas equation is $\text{n=}\frac{\text{PV}}{\text{RT}}$

Converting pressure from $\text{mmHg}$ into $\text{atm}$:

$\begin{array}{c}\text{19}\text{.2}\text{mmHg}=\text{19}\text{.2}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\because \text{1}\text{atm}=\text{760}\text{mmHg}\\ \text{=}\text{0}\text{.02526}\text{atm}\end{array}$

Substituting all the values in the ideal gas equation:

$\begin{array}{c}\text{=}\frac{\left(\text{0}\text{.02526}\text{atm}\right)\left(\text{0}\text{.843L}\right)}{\left(\text{0}\text{.0821}\text{L}\text{.}\text{atm/K}\text{.mol}\right)\left(\text{1235K}\right)}\\ \text{=}\text{2}{\text{.10×10}}^{\text{-4}}\text{mol}\end{array}$

So, the number of moles of the unknown alkali metal is $\text{2}{\text{.10×10}}^{\text{-4}}\text{mol}$

Calculating the volume/mol:

Given: Edge length is $\text{0}\text{.171}\text{cm}$.

The relation between volume and the edge length is $\text{V=}{\text{a}}^{\text{3}}$

$\begin{array}{c}\text{V}\text{=}{\left(\text{0}\text{.171}\text{cm}\right)}^3\\ =5.00\times {10}^{-3}{\text{cm}}^3\end{array}$

Thus, the volume of the body centered cubic unit cell is $5.00\times {10}^{-3}{\text{cm}}^3$ which is the volume of $\text{2}{\text{.10×10}}^{\text{-4}}\text{mol}$ unknown alkali metal.  So, it is simply the volume/mol.

Converting volume/mol into volume/unit cell:

$\begin{array}{l}\text{=}\frac{\text{5}\text{.00}{\text{×10}}^{\text{-3}}{\text{cm}}^{\text{3}}}{\text{2}{\text{.10×10}}^{\text{-4}}\text{mol}}\text{×}\frac{\text{1mol}}{\text{6}\text{.022}{\text{×10}}^{\text{23}}\text{atoms}}\text{×}\frac{\text{2}\text{atoms}}{\text{1}\text{unit}\text{cell}}\\ \text{=}\text{7}\text{.91}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\text{/unit}\text{cell}\end{array}$

Thus the volume of the unit cell is $\text{7}\text{.91}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\text{/unit}\text{cell}$.

Calculating the edge length with respect to this volume:

$\text{a}=\sqrt[3]{\text{V}}$

$\begin{array}{c}\text{a}=\sqrt[3]{\text{7}\text{.91}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}}\\ =4.2927\times {10}^{-8}\text{cm}\end{array}$

Thus the edge length is $\text{a}=4.2927\times {10}^{-8}\text{cm}$

It is known that all alkali metals have a body-centered cubic structure.

So, calculating the radius of the unknown alkali-metal in the body-centered cubic structure.

The radius of for a body-centered cubic unit cell can be expressed as follows:

$\text{r}\text{=}\frac{\sqrt{\text{3}}}{\text{4}}\times \text{a}$

$\begin{array}{c}=\frac{\sqrt{3}}{4}\times 4.2927\times {10}^{-8}\text{cm}\\ =1.8576\times {10}^{-8}\text{cm}\end{array}$

Converting radius from centimeters into picometers:

$\begin{array}{c}1.8576\times {10}^{-8}\text{cm}=1.8576\times {10}^{-8}\text{c}\bcancel{\text{m}}\text{×}\left(\frac{\text{1}{\text{×10}}^{\text{-2}}\bcancel{\text{m}}}{\text{1}\text{c}\bcancel{\text{m}}}\text{×}\frac{\text{1}\text{pm}}{\text{1}{\text{×10}}^{\text{-12}}\bcancel{\text{m}}}\right)\\ =1.8576\times {10}^{-8}{\text{×10}}^{\text{10}}\text{pm}\\ \text{=}1.8576{\text{×10}}^2\text{pm}\\ \text{=}185.76\text{pm}\end{array}$

Thus the radius is $185.76\text{pm}\approx 186\text{pm}$

Figure

This calculated radius is the radius of sodium metal.

So, the unknown metal is found to be sodium metal.

Calculating the mass of sodium metal:

Since sodium metal is in body-centered cubic lattice, hence there will be 2 atoms per unit cell.

The molar mass of sodium is 23 amu which is the mass of 1 sodium atom and the mass of $\text{6}\text{.023}\text{×}{\text{10}}^{\text{23}}\text{amu}$ is equal to 1g of sodium.

So, the 2 atoms of sodium per unit cell can be calculated as follows:

$\begin{array}{l}\text{2}\text{atoms}\text{×}\frac{\text{23}\text{amu}}{\text{1}\text{Na}\text{atom}}\text{×}\frac{\text{1g}}{\text{6}\text{.023}\text{×}{\text{10}}^{\text{23}}\text{amu}}\\ \text{=}\text{7}\text{.6374}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

So, the mass of $\text{2}\text{atoms}$ per unit cell is $\text{7}\text{.6374}\text{×}{\text{10}}^{\text{-23}}\text{g}$.

The volume of the unit cell has been calculated as $\text{7}\text{.91}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\text{/unit}\text{cell}$.

With all these calculations, calculating the density of sodium metal:

$\begin{array}{c}\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}\\ =\frac{\text{7}\text{.6374}\text{×}{\text{10}}^{\text{-23}}\text{g}}{\text{7}\text{.91}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\text{/unit}\text{cell}}\\ =0.9655\text{g}/{\text{cm}}^{\text{3}}\text{/unit}\text{cell}\approx 0.97\text{g}/{\text{cm}}^{\text{3}}\text{/unit}\text{cell}\end{array}$

The Actual density of sodium is known to be $0.97\text{g}$ which exactly matches with the calculated density of sodium. This proves the identification of the sodium metal.

Conclusion

From the calculation of radius, the unknown alkali metal has been identified and from the calculation of density, the identification has been proved.