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An element crystallizes in fec lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm^(-3). Calculate the number of atoms in 108 g of the element. |
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Answer» SOLUTION :`d=(zm)/(a^(3)N)`, m = Mass of element, N = number of atoms Or `N=(108xx4)/(10.8xx27xx10^(-24))=1.48xx10^(24)` atoms `M=(a^(3)xxN_(a)xxd)/(Z)` `=(27xx10^(-24)xx6.022xx10^(23)xx10.8)/(4)=43.88"g mol"^(-1)` `43.88"g mol"^(-1)` CONTAINS `6.02xx10^(23)` atoms So, 108 g contains`=(6.02xx10^(23)xx108)/(43.88)=1.48xx10^(24)` atoms |
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