An element crystallizes in b.c.c. lattice with cell edge of 500 pm. The density elements is 7.5 `gcm^-3`. How many atoms are present in 300 g of the element ?

Explanation

`rho =(ZM)/(a^(3) xx N_A)["Z for bcc"=2]`
`7.5 = (2 xx M)/((500 xx 10^(-10))^(3) xx 6.022 xx 10^(23))`
`M = (7.5 xx 500 xx 500 xx 500 xx 10^(-30) xx 6.022 xx 10^(23))/(2)`

M = 282.3 g/mol.
282.3 g of element contains `=6.022 xx 10^(23)` atoms.
300 g of element contains `=(6.022 xx 10^(23))/(282.3) xx 300 = 6.399 xx 10^(23)` atoms