Question

Silver forms ccp lattice and x-ray studies of its crystals show that theedge length of its unit cell is408.6 pm. Calculate the density of silver(atomi

Answer 1

Answer:

d-10.5gcm-3

Step-by-step explanation:

silver forms cubic close packing (ccp) , therefore the number of silver atoms per unit cell z=4

Molar mass of silver M=107.9gmol−1=107.9×10−3kgmol−1

Edge length of unit cell a=408.6pm=408.6×10−12

Density d=(z.M)/(a3.NA)

d=4×(107.9×10−3kgmol−1)/(408.6×10−12m)3(6.022×1023mol−1)

=10.5×103kgm−3

d=10.5gcm−3

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