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$6.02\times 10^{20}$ molecules of urea are present in 100ml solution.The concentration of urea solution is

$\begin{array}{1 1}(a)\;0.1 m&(b)\;0.01m\\(c)\;0.02m&(d)\;0.001m\end{array}$

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$M=\large\frac{Moles\;of\; urea}{V\;in\;litre}$
$\;\;\;=\large\frac{6.02\times 10^{20}\times 1000}{6.023\times 10^{23}\times 100}$
Hence (b) is the correct answer.
answered Oct 28, 2013 by sreemathi.v

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