Question

Tantalum (Ta;d=$\mathbf{16}\mathbf{.}\mathbf{634}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$and $\mathbf{M}\mathbf{=}\mathbf{180}\mathbf{.}\mathbf{9479}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}$) has a body-cantered cubic structure with a unit-cell edge length of 3.3058 Å. Use these data to calculate Avogadro’s number.

Short answer

The answer is Avogadro’snumber${\mathbf{N}}_{\mathbf{0}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{atoms}\mathbf{/}\mathbf{mol}$

Step-by-step solution

Concept

When a structure consists of one atom in the centre and each corner of the cube shares an atom arranged in a cube is called a body-centered cubic structure.

To calculate Avogadro’s number, we will use the following formula

$\mathrm{d}=\frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{N}}_0\times {\mathrm{a}}^3}$ …(1)

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{Molar}\mathrm{mass}\left(\mathrm{M}\right)\mathrm{of}\mathrm{Tantalum}=180.9479\mathrm{g}/\mathrm{mol} \\ \mathrm{Density}\left(\mathrm{d}\right)\mathrm{of}\mathrm{Tantalum}=16.634\mathrm{g}/{\mathrm{cm}}^3 \\ \mathrm{Since},\mathrm{structure}\mathrm{of}\mathrm{the}\mathrm{body}-\mathrm{centered}\mathrm{cubic}\mathrm{unit}\mathrm{cell} \\ \mathrm{Hence},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{atoms}\mathrm{per}\mathrm{unit}\mathrm{cell}=\left(\mathrm{Z}\right)=2\mathrm{atoms} \\ \mathrm{Edge}-\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{unit}\mathrm{of}\mathrm{cell}\left(\mathrm{a}\right)=3.3058\times {10}^{-8}\mathrm{cm} \\ \mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{Number}\left({\mathrm{N}}_0\right)=? \end{gathered}$$

Further Solution

By submitting the value in the eq. 1

$\begin{array}{rcl}{\mathrm{N}}_0& =& \frac{2\mathrm{atoms}\times 180.9479}{16.634\times {\left(3.3058\times {10}^{-8}\mathrm{cm}\right)}^3}\\ 16.634& =& \frac{2\mathrm{atoms}\times 180.9479}{{\mathrm{N}}_0\times {\left(3.3058\times {10}^{-8}\mathrm{cm}\right)}^3}\\ {\mathrm{N}}_0& =& \frac{361.8958}{16.634\times \left(36.1268192311\times {10}^{-24}\right)}\\ {\mathrm{N}}_0& =& \frac{361.8958}{600.933511090\times {10}^{-24}}\\ {\mathrm{N}}_0& =& .60222269738\times {10}^{24}\\ & & \end{array}$

Hence,data-custom-editor="chemistry" ${\mathrm{N}}_0=6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mol}$.