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(D) The formula for density $(ho)$ of a unit cell is given by: $ho = \frac{M 	imes n}{a^3 	imes N_A}$ Where $M$ is the molar mass,$n$ is the num

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$4$

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(D) The formula for density $(ho)$ of a unit cell is given by: $ho = \frac{M 	imes n}{a^3 	imes N_A}$ Where $M$ is the molar mass,$n$ is the number of atoms per unit cell,$a^3$ is the volume of the unit cell,and $N_A$ is the Avogadro constant. Given: $ho = 20 \ g \ cm^{-3}$,$M = 190 \ g \ mol^{-1}$,and $a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}$. Substituting the values into the formula: $20 = \frac{190 	imes n}{38}$ $n = \frac{20 	imes 38}{190}$ $n = \frac{760}{190} = 4$ Therefore,the number of atoms present in the unit cell is $4$.

Calculate the number of atoms present in a unit cell of an element having molar mass $190 \ g \ mol^{-1}$ and density $20 \ g \ cm^{-3}$. Given that $[a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}]$.

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