Question

silver forms ccp lattice and x-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. calculate the density of silve (At mass= 107.94)

Answer 1

I don't know plz check on Google

Answer 2

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

• 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 = ${a=408.6×10^-12m}$
• ${Density}$,${d}$=$\frac{z.M}{a^3.NA}$
• $\frac{4×(107.9×10^-3kgmol^-1)}{(408 6×10^-12m)^3(6.022×10^23mol^-1)}$ = ${10.5×10^3 kgm^-3}$
• ${10.5g cm^3}$.