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

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(C) The number of atoms in one body-centred cubic $(BCC)$ unit cell is:$(1/8 	imes 8) + 1 = 2 	ext{ atoms}$.The number of moles of potassium in $39

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The number of atoms in one body-centred cubic $(BCC)$ unit cell is:$(1/8 	imes 8) + 1 = 2 	ext{ atoms}$.The number of moles of potassium in $39 \ g$ is:$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{39 \ g}{39 \ g/mol} = 1 	ext{ mol}$.The total number of atoms in $39 \ g$ of potassium is:$1 	imes 6.023 	imes 10^{23} = 6.023 	imes 10^{23} 	ext{ atoms}$.Since each $BCC$ unit cell contains $2$ atoms,the number of unit cells is:$\frac{6.023 	imes 10^{23}}{2} = 3.011 	imes 10^{23} 	ext{ unit cells}$.

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

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