Question

An element with density $11.2$ $gm$ ${cm}^{-3}$ forms a fcc lattice with edge length of $4\times {10}^{-8}cm$. Calculate the atomic mass of the element. ($N_A=6.02\times {10}^{23}{mol}^{-1}$)

Answer 1

Given that,

Density, $d=11.2g/c{m}^3$
Edge length, $a=4\times {10}^{-8}cm$
Avogadro's number, $N_A=6.02\times {10}^{23}/mol$
Atomic mass $,M=??$

For f.c.c lattice, number of atoms per unit cell, $z=4$

Using the formula,

$d=\frac{Mz}{N_A{a}^3}$

$M=\frac{d{N}_A{a}^3}{z}$

$M=\frac{11.2\times 6.023\times {10}^{23}\times (4\times {10}^{-8}{)}^3}{4}$

$M=107.9\approx 108g/mol$

Hence, the atomic mass of the element is $108g/mol$. It is silver metal.