Numerical

An element crystallizes in fcc 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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#### Solution

**Given: **Mass of the element, W = 108 g

z (fcc) = 4

Edge length, (a) = 300 pm ...(1pm = 10−10 cm)

Density, (ρ) = 10.8 g/cm3

`"p" = (z xx "M")/(a^3 xx "N"_"A")`

`10.8 = (4 xx "M")/((300 xx 10^-10)^3 xx6.022 xx 10^23)`

`"M" = (10.8 xx 2.7 xx 10^-23 xx 6.022 xx 10^23)/(4) "g"`

`n = "N"/"N"_("A") = "W"/"M"`

`"N" = "W"/"M" xx "N"_"A"`

`"N" = (108 xx 4 xx 6.022 xx 10^23)/(10.8 x 2.7 xx 10^-23 xx 6.022 xx 10^23)`

`"N" = 14.8 xx 10^23 "atoms"`

Concept: Calculations Involving Unit Cell Dimensions

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