An element with a molar mass 27 text{ g mol}^ | vedclass

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(C) The density of a unit cell is given by the formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$Where $d$ is density, $Z$ is the number of atoms per un

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$4$

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(C) The density of a unit cell is given by the formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$Where $d$ is density, $Z$ is the number of atoms per unit cell, $M$ is molar mass, $a^3$ is the volume of the unit cell, and $N_A$ is Avogadro's number.Given: $d = 2.7 	ext{ g cm}^{-3}$, $M = 27 	ext{ g mol}^{-1}$, and $a^3 	imes N_A = 40 	ext{ cm}^3 	ext{ mol}^{-1}$.Rearranging the formula to solve for $Z$: $Z = \frac{d 	imes a^3 	imes N_A}{M}$Substituting the values: $Z = \frac{2.7 	imes 40}{27}$$Z = \frac{108}{27} = 4$Thus, the number of atoms present in the unit cell is $4$.

An element with a molar mass $27 \text{ g mol}^{-1}$ forms a cubic unit cell. Calculate the number of atoms present in a unit cell if the density of the metal is $2.7 \text{ g cm}^{-3}$. Given: $[a^3 \times N_A = 40 \text{ cm}^3 \text{ mol}^{-1}]$

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