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Calculate molecules of methane,C and H atoms in 25g methane

$\begin{array}{1 1}(a)\;9.41\times 10^{23}atom\;of\;C,3.764\times 10^{24}atoms\;of\;H\\(b)\;8.9\times 10^{23}atoms\;of\;C,3.764\times 10^{24}atoms\;of\;H\\(c)\;8.9\times 10^{23}atoms\;of\;C,4.2\times 10^{23}\;atoms\;of\;H\\(d)\;6.5\times 10^{23}atom\;of\;C,4.5\times 10^{23}atoms\;of\;H\end{array}$

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Mol.weight of $CH_4=16$
$16g\;CH_4=N$molecules
$25g\;CH_4=\large\frac{6.023\times 10^{23}\times 25}{16}$molecules
$\Rightarrow 9.41\times 10^{23}$ molecules
$16gCH_4=N\;atom\;of\;C$
$25g\;CH_4=\large\frac{6.023\times 10^{23}\times 25}{16}$atoms of C
$\Rightarrow 9.41\times 10^{23}$ atoms of C
Also,
$16gCH_4=N$atoms of H
$25gCH_4=\large\frac{4\times 25\times 6.023\times 10^{23}}{16}$atoms of H
$\Rightarrow 3.764\times 10^{24}$atoms of H
Hence (a) is the correct answer.
answered Oct 30, 2013 by sreemathi.v
 

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