Question

Silver forms ccp lattice and X-ray of studies its crystals show that the edge length of its unit cell 408.6pm. Calculate the density of silver (Atomic mass = 107.9 g/mol, NA =6.022 X 1023 mol)​

Answer 1

${\huge{\underline{\underline{\mathcal{\red{♡ANSWER♡}}}}}}$

• Since the lattice is ${CCP}$,the number of silver atoms per unit cell = ${z}$=${4}$
• Molar mass of silver = ${107.9mol^-1}$=${=107.9×10^-3 kg mol^-1}$
• Edge length of unit cell = ${a=408.6×10^-12m}$
• ${Density}$,${d}$=$\frac{z.M}{a^3.NA}$
• $\frac{4×(107.9×10^-3kgmol^-1)}{(408 6×10^-12m)^3(6.022×10^23mol^-1)}$ = ${10.5×10^3 kgm^-3}$
• ${10.5g cm^3}$.