Q12.101P - Chemistry: Molecular Nature Of Matter And Change

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.

Step-by-Step Solution

Answer

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

Step 1: 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" $$\begin{gathered} \mathrm{Molar} \mathrm{mass} \left(\mathrm{M}\right) \mathrm{of} \mathrm{Tantalum}=180.9479 \mathrm{g}/\mathrm{mol} \\ \mathrm{Density} \left(\mathrm{d}\right) \mathrm{of} \mathrm{Tantalum}=16.634 \mathrm{g}/{\mathrm{cm}}^3 \\ \mathrm{Since}, \mathrm{structure} \mathrm{of} \mathrm{the} \mathrm{body}-\mathrm{centered} \mathrm{cubic} \mathrm{unit} \mathrm{cell} \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{atoms} \mathrm{per} \mathrm{unit} \mathrm{cell}=\left(\mathrm{Z}\right)=2 \mathrm{atoms} \\ \mathrm{Edge}-\mathrm{length} \mathrm{of} \mathrm{the} \mathrm{unit} \mathrm{of} \mathrm{cell}\left(\mathrm{a}\right)=3.3058\times {10}^{-8} \mathrm{cm} \\ \mathrm{Avogadro}\text{'}\mathrm{s} \mathrm{Number}\left({\mathrm{N}}_0\right)=? \end{gathered}$$

Step2: 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" ${\mathrm{N}}_0=6.023\times {10}^{23} \mathrm{atoms}/\mathrm{mol}$.