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Total number of atoms present in $1.0cm^3$ of solid urea (density 0.3g/$cm^3$) at $25^{\large\circ}$ are

$\begin{array}{1 1}(a)\;3.01\times 10^{21}&(b)\;2.41\times 10^{22}\\(c)\;3.01\times 10^{22}&(d)\;2.41\times 10^{23}\end{array}$

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$W_{urea}=1\times 0.3g$
Also $60g$ urea has 8N atoms ($NH_2CONH_2)$
Hence (b) is the correct answer.
answered Oct 28, 2013 by sreemathi.v
 

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