An element with molar mass 2.7 times 10^{-2} | vedclass

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

Vedclass · Accepted Answer

$4$

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(B) The density of a unit cell is given by the formula: $ho = \frac{Z 	imes M}{N_A 	imes a^3}$ Where $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a$ is the edge length. Rearranging for $Z$: $Z = \frac{ho 	imes N_A 	imes a^3}{M}$ Given: $ho = 2.7 	imes 10^3 \ kg \ m^{-3}$ $M = 2.7 	imes 10^{-2} \ kg \ mol^{-1}$ $a = 405 \ pm = 405 	imes 10^{-12} \ m = 4.05 	imes 10^{-10} \ m$ $N_A = 6.023 	imes 10^{23} \ mol^{-1}$ Substituting the values: $Z = \frac{(2.7 	imes 10^3) 	imes (6.023 	imes 10^{23}) 	imes (4.05 	imes 10^{-10})^3}{2.7 	imes 10^{-2}}$ $Z = \frac{2.7 	imes 10^3 	imes 6.023 	imes 10^{23} 	imes 66.43 	imes 10^{-30}}{2.7 	imes 10^{-2}}$ $Z = \frac{1080.14 	imes 10^{-4}}{2.7 	imes 10^{-2}} \approx 4$ Thus, the number of atoms per unit cell is $4$.

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the number of atoms present in one unit cell of it is (Given : $N_{A}=6.023 \times 10^{23} \ mol^{-1}$)

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