An element has a bcc structure with a 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?

Explanation

Volume of unit cell `= (288 "pm")^(3) = (288 xx 10^(-12) m)^(3)`
`= (288 xx 10^(-10)cm^(3))`
`= 2.39xx 10^(-23) cm^(3)`
For bc c, `Z_(eff) = 2`
`:. rho = (Z_(eff) xx AW)/(N_(A) xx a^(3))`
`7.2 g cm^(-3) = (2 xx AW)/(6 xx 10^(23) xx 2.93 xx 10^(-23) cm^(3))`
`:. Aw = 7.2 xx 3 xx 10^(23) xx 2.39 xx 10^(-23) g`
`= 7.2 xx 3 xx 2.39 g`
`:. 7.2 xx 3 xx 2.93 g` of elements contains `= N_(A)` atoms
`= 6 xx 10^(23)` atoms.
`:. 208 g` of the element contains `= (6 xx 10^(23) xx 208)/(7.2 xx 3 xx 2.39)`
`= 24.17 xx 10^(23)` atoms
Alternatively
Volume of `208 g` of the element `= (Mass)/(Density) = (208 g)/(7.2 g cm ^(-3))`
`= 28.88 cm^(3)`
Number of unit cells in this volume
`= (28.88 cm^(3))/(2.39 xx 10^(23)) cm^(3)//"unit cell"`
`= 12.08 xx 10^(23)` unit cells
Since each cubic unit cell contains `2` atoms, therefore the totl number of atoms in `208 g`
`= 2 ("atom"//"unit cell") xx 12.08 xx 10^(23)` unit cells
`= 24.16 xx 10^(23)` atoms