A compound forms a bcc unit cell with an edge | vedclass

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(B) Step $1$: Identify the given parameters. For a $bcc$ unit cell, the number of atoms per unit cell $Z = 2$. Edge length $a = 400 	ext{ pm} = 400 \

Vedclass · Accepted Answer

$67.4 	ext{ g mol}^{-1}$

Vedclass · Answer

(B) Step $1$: Identify the given parameters. For a $bcc$ unit cell, the number of atoms per unit cell $Z = 2$. Edge length $a = 400 	ext{ pm} = 400 	imes 10^{-10} 	ext{ cm} = 4 	imes 10^{-8} 	ext{ cm}$. Density $d = 3.5 	ext{ g cm}^{-3}$. Avogadro's number $N_A = 6.022 	imes 10^{23} 	ext{ mol}^{-1}$.Step $2$: Use the density formula for a unit cell: $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Step $3$: Rearrange the formula to solve for molar mass $M$: $M = \frac{d 	imes a^3 	imes N_A}{Z}$.Step $4$: Substitute the values: $M = \frac{3.5 	imes (4 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{2}$.Step $5$: Calculate $a^3 = 64 	imes 10^{-24} 	ext{ cm}^3$. $M = \frac{3.5 	imes 64 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{2} = \frac{3.5 	imes 64 	imes 0.6022}{2} = 3.5 	imes 32 	imes 0.6022 = 67.4464 	ext{ g mol}^{-1}$.Thus, the molar mass is approximately $67.4 	ext{ g mol}^{-1}$.

$A$ compound forms a $bcc$ unit cell with an edge length of $400 \text{ pm}$. Determine the molar mass of the compound if the density of the compound is $3.5 \text{ g cm}^{-3}$. (Given: $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$)

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