The Solid State

Question

An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (N_{A} = 6.022 × 1023 mol^{–1} )

Answer

We know,

$d=\frac{ZxM}{{N}_{A}x{a}^{3}}\phantom{\rule{0ex}{0ex}}ForFCCZis4\phantom{\rule{0ex}{0ex}}d=\frac{4x40}{6.022x{10}^{23}x(400x{10}^{-10}{)}^{3}}=4.1514gmc{m}^{-3}\phantom{\rule{0ex}{0ex}}Volumeof4gmXis\frac{4}{4.1514}c{m}^{3}=0.96c{m}^{3}\phantom{\rule{0ex}{0ex}}Volumeof1unitcellis=(400x{10}^{-10}{)}^{3}c{m}^{3}=64x{10}^{-24}c{m}^{3}\phantom{\rule{0ex}{0ex}}Numberofunitcell=\frac{0.96}{64x{10}^{-24}}=1.5x{10}^{22}$