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 ?

An element has a body-centred cubic (bcc) structure with cell edge of

1.	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» Solution :For the BCC structure, Z=2 EDGE of the unit cell , a=288 pm , DENSITY of the element , `rho=7.2 g//cm^3` Subtituting the values in the expression `rho=(ZxxM)/(a^3xxN_0)` `7.2 g cm^(-3) =(2xxM)/((288xx10^(-10) cm)^3 XX(6.02xx10^23 mol^(-1))) "or" M=51.8 g mol^(-1)` By mole concept, 51.8 g of the element contains =`6.02xx10^23` atoms `therefore` 208 g of the element contains =`(6.02xx10^23)/51.8xx208` atoms =`24.17xx10^23` atoms