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An element exists in the body-centred cubic structure whose cell edge is 2.88 A. The density of the element is 7.20 g/cm^2. Calculate the number of atoms in 104 g of the element. |
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Answer» Solution :Volume of unit cell `=(2.88 XX 10^(-8))^(3) cc = 2.39 xx 10^(-23)` cc Volume of the element weighing 104 g = `("MASS")/("density")` `therefore` number of unit cells present in 104 g of the element `=(14.44)/(2.39 xx 10^(-23)) = 6.04 xx 10^(23)` Since each body-centred CUBIC cell contains 2 atoms, number of atoms `=2 xx 6.04 xx 10^(23) = 1.208 xx 10^(24)` |
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