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An element crystallizes in a f.c.c. lattice with cell edge of $400 \; pm.$ The density of the element is $7\; g\; cm^{-3}.$ How many atoms are present in $280\; g$ of the element ?

Answers (1)

$Edge (a) =400 pm\; \; \; \; mfcc,z=4$

$Density =7g/cm^{3}$

$mass=280\; g$

$No.\; of\; atoms=?$

Volume = $(a)^{3}=(400pm)^{3}$

$=6.4\times 10^{-23}cm^{3}$

Volume of $280\; g=\frac{mass}{density}$

$=\frac{280}{7}=40\; cm^{3}$

No. of unit cells in $400cm^{3}=\frac{40}{6.4\times 10^{-23}}$

$=6.25\times 10^{23}unit \; cells$

Total no. of atoms in $280\; g=4\times 6.25\times 10^{23}$

$=25\times 10^{23}atoms$

Posted by

Sumit Saini

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