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In the standardization of $Nas_2S_2O_3$ using $K_2Cr_2O_7$ by iodometry , the equivalent weight of $K_2Cr_2O_7$ is

$(a)\;\text{(molecular weight)/2} \\ (b)\;\text{(molecular weight )/6} \\(c)\;\text{(molecular weight)/3} \\(d)\;\text{Same as molecular weight} $
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In iodometry , $K_2Cr_2O_7$ acts as an oxidising agent to liberate iodine $(I_2)$ from an iodide such as $KI$ or $NaI$
The iodine $(I_2) $ thus liberated. is titrated with hypo $(Na_2S_2O_3)$ solution.
$Na_2S_2O_3+I_2 \longrightarrow Na_2S_4O_6 +2 NaI$
We find that one mole of $K_2Cr_2O_7$ accepts 6 moles of electron as follows.
$Cr_2O_7^{2-}+14H^{+}+6e^{-} \longrightarrow 2Cr{3+}+7H_2O$
Thus equivalent weight $=\large\frac{Molecular \;weight}{6}$
Hence b is the correct answer.
answered Mar 21, 2014 by meena.p

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