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From 200mg of $CO_2,10^{21}$ molecules are removed.How many moles of $CO_2$ are left?

$\begin{array}{1 1}(a)\;3.1\times 10^{-3}&(b)\;2.4\times 10^2\\(c)\;2.88\times 10^{-3}&(d)\;2.88\times 10^{-7}\end{array}$

1 Answer

Grams-molecular mass of $CO_2=44g$
Mass of $10^{21}$ molecules of $CO_2=\large\frac{44}{6.02\times 10^{23}}$$\times 10^{21}$
$\Rightarrow 0.073g$
Mass of $CO_2$ left=$(0.2-0.073)=0.127g$
Number of moles of $CO_2$ left=$\large\frac{0.127}{44}$
$\qquad\qquad\qquad\qquad\qquad=2.88\times 10^{-3}$
answered Oct 22, 2013 by sreemathi.v
 

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