Question

An element has a body-centred cubic (bcc) structure with cell edge of 288 pm. The density of the element is 7.2 g cm-3. How many atoms are present in 208 g of the element?

Answer 1

For ‘bcc’ Z = 2
Edge length ‘a’ = 288 pm = 288 x 10-10 cm
Density (d) = 7.2 g cm-3, Mass of element = 208 g

= 51.8 g mol-1
According to mol concept 51.8 g element has = 6.022 x 1023atoms
∴208 g of element will have = (\frac { 208 }{ 51.8 } \times 6.022) x 1023 atoms
= 24.17 x 1023 atoms