Question

An element, density 6.8 g cm-3occurs in bcc structure with cell edge 290 pm. Calculate the number of atoms present in 200 g of the element.​

Answer 1

Answer:

Explanation:

Here is the answer

Answer 2

Number of atoms present in 200 g of the element are $2.41 \times 10^{24}$.

Explanation:

The given data is as follows.

         For bcc structure, Z = $\frac{1}{8} \times 8 + 1$ = 2

         Density = 6.8 $g/cm^{3}$

         mass (M) = 200 g

         Edge (a) = 290 pm = $290 \times 10^{-12} m$ = $290 \times 10^{-10} cm$

Therefore, calculate the number of atoms as follows.

               d = $\frac{Z \times M}{N \times a^{3}}$

Putting the given values into the above formula as follows.

           d = $\frac{Z \times M}{N \times a^{3}}$

       6.8 $g/cm^{3}$ = $\frac{2 \times 200 g}{N \times (290 \times 10^{-10}^{3})}$

                  N = $2.41 \times 10^{24}$

Therefore, number of atoms present in 200 g of the element are $2.41 \times 10^{24}$.

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