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Question

(B) $Ag$ crystallizes in a face-centered cubic $(FCC)$ lattice.In an $FCC$ lattice,the distance between two nearest atoms is $d = \frac{a}{\sqrt{2}}$,

Vedclass · Accepted Answer

$16.8$

Vedclass · Answer

(B) $Ag$ crystallizes in a face-centered cubic $(FCC)$ lattice.In an $FCC$ lattice,the distance between two nearest atoms is $d = \frac{a}{\sqrt{2}}$,where $a$ is the edge length.Given $d = 2.5 	imes 10^{-8} \ cm$,so $a = d 	imes \sqrt{2} = 2.5 	imes 10^{-8} 	imes 1.414 = 3.535 	imes 10^{-8} \ cm$.For $FCC$,the number of atoms per unit cell $Z = 4$.The density formula is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Substituting the values: $d = \frac{4 	imes 108}{6.022 	imes 10^{23} 	imes (3.535 	imes 10^{-8})^3}$.$d = \frac{432}{6.022 	imes 10^{23} 	imes 44.17 	imes 10^{-24}} = \frac{432}{26.6} \approx 16.24 \ g/cm^3$.Rounding to the nearest provided option,the correct answer is $16.8 \ g/cm^3$.

The distance between two nearest $Ag$ atoms is $2.5 \times 10^{-8} \ cm$. The density of its crystal is .......... $g/cm^3$. $(Ag = 108 \ g/mol)$

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