Question

silver forms CCP lattice and x-ray study of its crystal show that the edge length of its unit cell is 40 8.6 picometre calculate the density of silver

Answer 1

CCP mean FCC face centred cubic metallic crystal
no . of atoms per unit cell in FCC = 4
given edge length (a) = 408.6 picometre
= 408.6*10^-10 cm
atomic mass of silver = 108 gm
since density
=no. of atoms per unit length*(atomic mass)/Na*(a)^3
so = 4*108/(6.023*10^23)* (408.6)^3*10^-30
= 432/6.023*68217388.1*10^-7
= 432*10^7/410873328 = 10.5 gm/cm^3(approx)

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