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Silver (atomic weight = $107.88$) crystallises with f.c.c lattice for which the side length of the unit cell is $4.0774\overset{\circ}{A}$. Density of $Ag$ is $10.53\; g\;cm^{-3}$. Calculate the Avogadro's number.

$(a)\;6.023\times10^{23}\qquad(b)\;6.045\times10^{23}\qquad(c)\;6.045\times10^{22}\qquad(d)\;6.023\times10^{22}$

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Avogadro No. = $\large\frac{n\times at.wt}{density\times a^3}$
$=\large\frac{4\times107.88}{10.53\times(4.0774\times10^{-8})^3}$
$=6.045\times10^{23}$
Hence answer is (b)
answered Feb 27, 2014 by sharmaaparna1
 

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