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InterviewSolution
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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» For f.c.c structure z =4 `a=300 "pm"=300x10^(-10)cm,d=10.8cm^(-3)` `d=(zxxM)/(a^(3)xxNA)` `M=(dxxa^(3)xxNA)/z=(10.8xx(300xx10^(-10)^(3)xx6.022xx10^(23)))/(4)=43.9 gmol^(-1)` `because` 43.9 g of element contains = `6.022xx10^(23)` atoms 108 g of the elements contains `=(6.022xx10^(23))/43.9xx108=14.81xx10^(23)` atoms |
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