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Calculate the number of atoms of oxygen present in 88g $CO_2$.What would be the weight of CO having the same no of oxygen atoms ?

$\begin{array}{1 1}(a)\;23.1\times 10^{23}O_2,110g\;CO\\(b)\;24.092\times 10^{23}O_2,112g\;CO\\(c)\;22.03\times 10^{22}O_2,115g\;CO\\(d)\;1.11\times 10^{20}O_2,120g\;CO\end{array}$

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Molecular weight of $CO_2$ =44 and it has 32g $O_2$ and one molecule of $O_2$ has 2 atoms.
$44g\;CO_2$ has $2N$ atoms of O
$88g$ $CO_2$ has 2N atoms =$\large\frac{2\times 6.023\times 10^{23}\times 88}{44}$
$\Rightarrow 24.092\times 10^{23}$atoms of $O_2$
N atoms of O present in 28g CO
$24.092\times 10^{23}$ atoms of O=$\large\frac{28\times 24.092\times 10^{23}}{6.023\times 10^{23}}$
$\Rightarrow 112g$ CO
Hence (b) is the correct option.
answered Oct 30, 2013 by sreemathi.v

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