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4.4g of $CO_2$ and $2.24$ litre of $H_2$ at STP are mixed in a container.The total number of molecules present in the container will be

$\begin{array}{1 1}(a)\;6.022\times 10^{23}&(b)\;1.2044\times 10^{23}\\(c)\;2\;mole&(d)\;6.023\times 10^{24}\end{array}$

1 Answer

44g $CO_2=N$molecules
$N=6.023\times 10^{23}$
$\therefore 4.4g CO_2=\large\frac{N}{44}$$\times 4.4$
$\Rightarrow \large\frac{N}{10}$
22.4 litres $H_2$ at STP=N molecules.
$\therefore$ 2.24 litres $H_2$ at STP$=\large\frac{N}{10}$molecules
Thus total molecule =$\large\frac{N}{10}+\frac{N}{10}$
$\qquad\qquad\quad\quad\quad=\large\frac{N}{5}$
(i.e) $\Rightarrow \large\frac{6.023\times 10^{23}}{5}$
$\Rightarrow 1.2044\times 10^{23}$
answered Oct 24, 2013 by sreemathi.v
 

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