1.

Anelement has a body-centred cubic (bec) 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 :According to the given data,
Number of ATOMS in one unit cell, Z= 2 (for bcc unit cell)
EDGE length, a= 288 pm = `288 XX 10^(-10)` CM
Density of element, d= `7.2 g cm^(-3)`
Substituting the values in the relation, `d = (Z xx M)/(N_A xx a^3)`
`7.2 g cm^(-3) = (2xx M)/((6.022 xx 10^(23) mol^(-1)) xx (288 xx 10^(-10) cm)^3)` or M=51.8 g `mol^(-1)`
By mole concept, 51.8 g of the element contains atoms = `6.022 xx 10^(23)`
`:.` 208 g of the element contains atoms = `(6.022 xx 10^(23))/(51.8) xx 208`
= `2.418 xx 10^(24)`


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