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An element has a face centered cubic unit cell with a length of 352.4 pm along an edge. The density of the element is "8.9 gcm-3". How many atoms are present in 100 g of an element ? |
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Answer» Solution :`"MASS "=100g` `"Density"="8.9 g CM"^(-3)` `"Edge length"="352.4 pm"` `(a)=352.4xx10^(-10)cm` `"Volume of the unit cell,"` `a^(3)=(352.4xx10^(-10)cm)=4.37xx10^(-23)cm^(3)` `"Volume of 100 g of an element,"` `=("Mass")/("Density")` `=(100)/(8.9)cm^(3)=11.23cm^(3)` Therefore number of unit CELLS, `=(11.23)/(4.37xx10^(-23))=2.56xx10^(23)` SINCE each Fcc cube contains 4 ATOMS, therefore total number of atoms in 100 g. `=4xx(2.56xx10^(23))=10.24xx10^(23)" atoms"` |
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