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Number of ions present in 2.0 litre of a solution of $0.8MK_4Fe(CN)_6$ is

$\begin{array}{1 1}(a)\;4.8\times 10^{22}&(b)\;4.8\times 10^{24}\\(c)\;9.6\times 10^{24}&(d)\;9.6\times 10^{22}\end{array}$

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1 Answer

Mole of $K_4Fe(CN)_6=2\times 0.8$
$\Rightarrow 1.6$
Also 1 mole of $K_4Fe(CN)_6$ gives $4K^+$ and $1Fe(CN)_6^{4-}$ ion
Thus total ions in 1.6 mole=$1.6\times 5\times N$
$\Rightarrow 48.184\times 10^{23}$ ions.
Hence (b) is the correct answer.
answered Oct 25, 2013 by sreemathi.v
 

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