Question

Silver crystallizes in fcc structure with edge length of unit cell, $408.7pm$. Hence density of silver is $(Ag=108gmo{l}^{-1})$

• A. $10.5gc{m}^{-3}$
• B. $9.82gc{m}^{-3}$
• C. $12.06gc{m}^{-3}$
• D. $8.86gc{m}^{-3}$

Answer 1

Answer: (A)

$10.5gc{m}^{-3}$

Density $\rho =\frac{Z\times M}{N_0\times a^3}$
$Z=4atoms=$ number of atoms in fcc unit cell
$M=108g/mol=$ atomic mass of silver
$N_0=6.023\times {10}^{23}atoms/mol=$ Avogadro's number
$a=408.7pm=408.7\times {10}^{-10}cm=$ edge length of unit cell

$\rho =\frac{Z\times M}{N_0\times a^3}$

$\rho =\frac{4atoms\times 108g/mol}{6.023\times {10}^{23}atoms/mol\times (408.7\times {10}^{-10}cm{)}^3}$

$\rho =10.5g/c{m}^3$