The number of atoms in 4.5 g of a face-centr | vedclass

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(D) For a face-centred cubic $(FCC)$ crystal, the number of atoms per unit cell is $Z = 4$. Edge length $a = 300 \ pm = 300 	imes 10^{-10} \ cm = 3 \

Vedclass · Accepted Answer

$6.6 	imes 10^{22}$

Vedclass · Answer

(D) For a face-centred cubic $(FCC)$ crystal, the number of atoms per unit cell is $Z = 4$. Edge length $a = 300 \ pm = 300 	imes 10^{-10} \ cm = 3 	imes 10^{-8} \ cm$. Density $d = 10 \ g \ cm^{-3}$. Using the density formula: $d = \frac{Z 	imes M}{N_A 	imes a^3}$, where $M$ is the molar mass. $M = \frac{d 	imes N_A 	imes a^3}{Z} = \frac{10 	imes 6.022 	imes 10^{23} 	imes (3 	imes 10^{-8})^3}{4}$. $M = \frac{10 	imes 6.022 	imes 10^{23} 	imes 27 	imes 10^{-24}}{4} = \frac{162.594}{4} = 40.6485 \ g \ mol^{-1}$. Now, the number of atoms in $4.5 \ g$ is calculated as: $	ext{Number of atoms} = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{4.5}{40.6485} 	imes 6.022 	imes 10^{23} \approx 6.66 	imes 10^{22}$ atoms.

The number of atoms in $4.5 \ g$ of a face-centred cubic crystal with edge length $300 \ pm$ is (Given: Density $= 10 \ g \ cm^{-3}$ and $N_A = 6.022 \times 10^{23}$)

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