The number of unit cells present in 39 g of potassium if it crystallizes as body centred cube is ( N = Avogadro number, At. Wt. of potassium = 39 g/mol )

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Potassium has bcc system
Number of mole $= 39/39 = 1$ mole
$1$ mole of atoms = $6 .022 \times 10^{23}$ atoms $= N$
$∵$ In bcc system $2$ atoms are present in $1$ unit cell.

$∴ N$ number of atoms are present in $\frac{N}{2}$ unit cells.

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