Question

an element has a body centred cubic (bcc) structure with a cell edge of 288pm. the density of the element is 7.2 g/cm How many atoms are present in 208g of the element?

Answer 1

Answer on Question#39740-Chemistry-Inorganic Chemistry

Question

An element has a body centred cubic (bcc) structure with a cell edge of $288~\mathrm{pm}$ . The density of the element is $7.2~\mathrm{g/cm^3}$ . How many atoms are present in $208~\mathrm{g}$ of the element?

Solution

Volume of given mass of the element:

$$V = \frac {m}{\rho} = \frac {2 0 8 g}{7 . 2 g / c m ^ {3}} = 2 9 c m ^ {3} = 2. 9 \cdot 1 0 ^ {- 5} m ^ {3}$$

Volume of an elementary cell ( $a$ -cell edge):

$$V _ {c e l l} = a ^ {3} = (2 8 8 p m) ^ {3} = (2 8 8 \cdot 1 0 ^ {- 1 2} m) ^ {3} = 2. 4 \cdot 1 0 ^ {- 2 9} m ^ {3}$$

Number of the cells in given mass of the element:

$$n _ {c e l l s} = \frac {V}{V _ {c e l l}} = \frac {2 . 9 \cdot 1 0 ^ {- 5} m ^ {3}}{2 . 4 \cdot 1 0 ^ {- 2 9} m ^ {3}} = 1. 2 \cdot 1 0 ^ {2 4} c e l l s$$

In case of body centred structure each elementary cell contains 2 atoms: one in the cell centre and 8 eighth parts $(1 + 8 \cdot 1/8 = 2)$ .

So, the number of atoms is two times greater than the number of cells:

$$n _ {a t o m s} = 2 \cdot n _ {c e l l s} = 2 \cdot 1. 2 \cdot 1 0 ^ {2 4} c e l l s = 2. 4 \cdot 1 0 ^ {2 4} a t o m s$$

Answer: $2.4 \cdot 10^{24}$ atoms