1.

An element has a body-centred cubic (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 280 g of the element ?

Answer»

Solution :VOLUME of UNIT cell `= a^3 xx 10^(-30) "cm"^3`
`= (288)^3 xx 10^(-30) "cm"^3`
No. of atoms PER unit cell (Z) = 2
`therefore M = (d xx N_A xx a^3 xx10^(-30))/Z`
`therefore M = (7.2 xx 6.022 xx 10^(23) xx (288)^3 xx 10^(-30))/(2)`
`therefore`M = 51.786 g/mol
`therefore`Number of atoms in 208 g element
`= 208/51.786 xx 6.022 xx 10^(23) = 24.18 xx 10^23` atoms.


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