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

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(C) Number of moles of $K = \frac{39 \ g}{39 \ g/mol} = 1 \ mol$. Number of $K$ atoms $= 1 \ mol 	imes 6.022 	imes 10^{23} \ atoms/mol = 6.022 	ime

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$\frac{6.02 	imes 10^{23}}{2}$

Vedclass · Answer

(C) Number of moles of $K = \frac{39 \ g}{39 \ g/mol} = 1 \ mol$. Number of $K$ atoms $= 1 \ mol 	imes 6.022 	imes 10^{23} \ atoms/mol = 6.022 	imes 10^{23} \ atoms$. Since potassium crystallises in a body-centred cubic $(bcc)$ structure,the number of atoms per unit cell is $2$. Therefore,the number of unit cells $= \frac{	ext{Total number of atoms}}{	ext{Atoms per unit cell}} = \frac{6.022 	imes 10^{23}}{2}$.

How many unit cells are present in $39 \ g$ of potassium that crystallises in body-centred cubic structure? [At. wt. of $K = 39$]

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