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

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$4$

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(C) The density of a unit cell is given by the formula: $ho = \frac{Z 	imes M}{N_A 	imes V}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_A$ is Avogadro's number,and $V$ is the volume of the unit cell.Given that the product of density $(ho)$ and volume $(V)$ is $1.8 	imes 10^{-22} \ g$,we have $ho 	imes V = 1.8 	imes 10^{-22} \ g$.We know that the mass of a unit cell is equal to $ho 	imes V$.Also,the mass of a unit cell is equal to the number of atoms $(Z)$ multiplied by the mass of a single atom $(m_{atom})$.Therefore,$Z 	imes m_{atom} = ho 	imes V$.Substituting the given values: $Z 	imes (4.5 	imes 10^{-23} \ g) = 1.8 	imes 10^{-22} \ g$.$Z = \frac{1.8 	imes 10^{-22}}{4.5 	imes 10^{-23}} = \frac{18 	imes 10^{-23}}{4.5 	imes 10^{-23}} = 4$.Thus,the number of atoms per unit cell is $4$.

Calculate the number of atoms present per unit cell if the product of density and volume of the unit cell is $1.8 \times 10^{-22} \ g$. [Mass of an atom $= 4.5 \times 10^{-23} \ g$]

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