What is the equivalent weight of SnCl_{2} in | vedclass

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(A) The given reaction is: $SnCl_{2} + Cl_{2} \longrightarrow SnCl_{4}$.In this reaction,the oxidation state of $Sn$ changes from $+2$ in $SnCl_{2}$ t

Vedclass · Accepted Answer

$95$

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(A) The given reaction is: $SnCl_{2} + Cl_{2} \longrightarrow SnCl_{4}$.In this reaction,the oxidation state of $Sn$ changes from $+2$ in $SnCl_{2}$ to $+4$ in $SnCl_{4}$.The $n$-factor is the change in oxidation state per molecule,which is $|4 - 2| = 2$.The molar mass of $SnCl_{2} = 118.7 + 2 	imes 35.5 = 118.7 + 71 = 189.7 \approx 190 \ g/mol$.The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{190}{2} = 95$.

What is the equivalent weight of $SnCl_{2}$ in the following reaction: $SnCl_{2} + Cl_{2} \longrightarrow SnCl_{4}$?

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