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An element X with an atomic mass of 60 g/mol has density of 6.23 g cm^(-3). If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this element. |
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Answer» Solution :`M="60 g mol"^(-1), "a = 400 pm "=400xx10^(-10)cm, d="6.23 g cm"^(-3)," z needs to be CALCULATED."` `z=(d XXA^(3)xxN_(A))/(M)` Substituting the values, we have `z=(6.23xx400^(3)xx10^(-30)xx6.023xx10^(23))/(60)=(6.23xx64xx10^(-1)xx6.023)/(60)=4` In has face - CENTRED cubic (fcc) structure. `4r=sqrt2a or r=(sqrt2a)/(4)=(1.414xx4xx10^(-8))/(4)=1.414xx10^(-8)cm=141.4" pm"`. |
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