Silver has an atomic radius of 144 pm and th | vedclass

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$(A)$ The atomic mass of silver $(Ag)$ is $107.87 \ g \ mol^{-1}$.For an $FCC$ lattice, the relation between edge length $(a)$ and radius $(r)$ is $a

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It has $FCC$ crystal lattice

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$(A)$ The atomic mass of silver $(Ag)$ is $107.87 \ g \ mol^{-1}$.For an $FCC$ lattice, the relation between edge length $(a)$ and radius $(r)$ is $a = 2\sqrt{2}r$.$a = 2 	imes 1.414 	imes 144 \ pm = 407 \ pm = 4.07 	imes 10^{-8} \ cm$.The density $(ho)$ is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.For $FCC$, $Z = 4$. So, $ho = \frac{4 	imes 107.87}{6.022 	imes 10^{23} 	imes (4.07 	imes 10^{-8})^3} \approx 10.5 \ g \ cm^{-3}$.Since the calculated density matches the given density $(10.6 \ g \ cm^{-3})$, silver crystallizes in an $FCC$ lattice.In an $FCC$ lattice, the packing efficiency is $0.74$, meaning $0.26$ fraction of the volume is empty, not the edge.

Silver has an atomic radius of $144 \ pm$ and the density of silver is $10.6 \ g \ cm^{-3}$. Based on this information, identify the correct statement.

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