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Question
An element has 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.
Solution
An element has bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm−3. For the Bec structure, n = 2
ρ = `"nM"/("a"^3"N"_"A")`
7.2 g cm−3 = `(2"M")/((288 xx 10^-10 "cm")^3 xx (6.023 xx 10^23 "mol"^-1))`
7.2 g cm−3 = `(2"M")/((2.38 xx 10^-23 "cm"^3) xx (6.023 xx 10^23 "mol"^-1))`
7.2 g cm−3 = `(2"M")/(14.33 "cm"^3 "mol"^-1)`
7.2 g = 0.140 M mpl
M = `(7.2 "g")/(0.140 "mol")`
= 51.42 g mol−1
By mole concept, 51.42 g of the element contains 6.023 × 1023 atom 208 g of the element will contain
= `(6.023 xx 10^23 xx 208)/51.42` atoms
= 24.17 × 1023 atoms or 2.417 × 1024 atoms
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