What is the equivalent weight of $S n {C l}_{2}$ in the following reaction?

$S n {C l}_{2} + {C l}_{2} ⟶ S n {C l}_{4}$

95

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45

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60

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30

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Solution

The correct option is A $95$
$S n {C l}_{2} + {C l}_{2} ⟶ S n {C l}_{4}$

$n_{f}$ for $S n C l_{2} = + 2$ as its oxidation state changes from $+ 2$ to $+ 4$

Equivalent weight of $S n C l_{2} = \frac{molar mass}{n_{f}} = \frac{190}{2} = 95$

Hence, the correct option is $A$

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