When silver crystallizes, it forms face-cente | vedclass

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(B) $1$. For a face-centered cubic $(FCC)$ unit cell, the number of atoms per unit cell, $Z = 4$.$2$. The formula for density $(d)$ is $d = \frac{Z

Vedclass · Accepted Answer

$10.49$

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(B) $1$. For a face-centered cubic $(FCC)$ unit cell, the number of atoms per unit cell, $Z = 4$.$2$. The formula for density $(d)$ is $d = \frac{Z 	imes M}{a^3 	imes N_A}$, where $a^3$ is the volume of the unit cell.$3$. Given: $Z = 4$, $M = 108 	ext{ g/mol}$, $V = a^3 = 6.84 	imes 10^{-23} 	ext{ cm}^3$, $N_A = 6.022 	imes 10^{23} 	ext{ mol}^{-1}$.$4$. Substituting the values: $d = \frac{4 	imes 108}{6.84 	imes 10^{-23} 	imes 6.022 	imes 10^{23}}$.$5$. $d = \frac{432}{6.84 	imes 6.022} = \frac{432}{41.19} \approx 10.49 	ext{ g cm}^{-3}$.

When silver crystallizes, it forms face-centered cubic $(FCC)$ unit cells. If the volume of the unit cell is $6.84 \times 10^{-23} \text{ cm}^3$, calculate the density of silver. (Molar mass of silver is $108 \text{ g/mol}$, $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$) (in $\text{ g cm}^{-3}$)

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