How many unit cells are present in 39 g of p | vedclass

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(C) The number of atoms per unit cell in a $BCC$ structure is calculated as follows:$Z = (1/8 	imes 8) + (1 	imes 1) = 1 + 1 = 2 	ext{ atoms per un

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$\frac{6.023 	imes 10^{23}}{2}$

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(C) The number of atoms per unit cell in a $BCC$ structure is calculated as follows:$Z = (1/8 	imes 8) + (1 	imes 1) = 1 + 1 = 2 	ext{ atoms per unit cell}$.Next,calculate the number of moles of potassium in $39 \ g$:$	ext{Moles} = \frac{	ext{Given mass}}{	ext{Atomic mass}} = \frac{39 \ g}{39 \ g/mol} = 1 	ext{ mol}$.The total number of atoms in $39 \ g$ of potassium is:$	ext{Total atoms} = 1 	imes 6.023 	imes 10^{23} = 6.023 	imes 10^{23} 	ext{ atoms}$.Finally,the number of unit cells is given by:$	ext{Number of unit cells} = \frac{	ext{Total atoms}}{	ext{Atoms per unit cell}} = \frac{6.023 	imes 10^{23}}{2}$.

How many unit cells are present in $39 \ g$ of potassium that crystallizes in a body-centered cubic $(BCC)$ structure? [Atomic weight of $K = 39$]

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