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How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00g.

$\begin{array}{1 1}2.57\times 10^{21}\\3.57\times 10^{21}\\0.57\times 10^{21}\\1.57\times 10^{21}\end{array} $

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One unit cell of NaCl contains 4 NaCl units (ie) has mass
$\Rightarrow \large\frac{4\times 58.5}{6.023\times 10^{23}}$g
No. of unit cell in 1g =$\large\frac{6.02\times 10^{23}}{4\times 58.5}$
$\Rightarrow 2.57\times 10^{21}$
answered Aug 4, 2014 by sreemathi.v
 
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