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A silicon (atomic weight=28) chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. How many silicon (Si) atoms are present in this chip.

$\begin{array}{1 1}(a)\;2.02\times 10^{-4}&(b)\;1.22\times 10^{20}\\(c)\;12.2\times 10^{20}&(d)\;122\times 10^{20}\end{array}$

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Answer: $1.22\times 10^{20}$ atoms
Mass of silicon, $5.68\;mg=5.68\times 10^{-3}g$
Number of Si atoms =$N_A\times $ g-atom
$= 6.023\times 10^{23}\times \large\frac{5.68\times 10^{-3}}{28}$
$= 1.22\times 10^{20}$ atoms
answered Oct 25, 2013 by sreemathi.v
edited Mar 22, 2014 by sharmaaparna1

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