How much amount of CuSO 4 ⋅ 5 H 2 O is required for liberation of 2.54 g of I 2 when titrated with KI ?

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Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsTitrationMedium2 minPYQ_2011

ChemistryMediumsingle choice

How much amount of $CuSO_{4} \cdot 5 H_{2} O$ is required for liberation of $2.54 g$ of $I_{2}$ when titrated with $KI$ ?

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$2 CuSO_{4} \cdot 5 H_{2} O + 4 KI ⟶ Cu_{2} I_{2} + 2 K_{2} SO_{4} + I_{2} + 10 H_{2} O 254 g$ Molecular weight of $2 CuSO_{4} \cdot 5 H_{2} O$ $[2 (63.5 + 32 + 64) + 10 (18)] g = 499 g$ $254 g$ of $I_{2}$ is liberated by $499 g CuSO \cdot 5 H_{2} O$ $2.54 g$ of $I_{2}$ will be liberated by $x$ g CuSO $_{4} \cdot 5 H_{2} O$ $x = \frac{499}{254} \times 2.54 = 4.99 g$

Stream:AIIMSSubject:ChemistryTopic:Redox ReactionsSubtopic:Titration

⏱ 2mℹ️ Source: PYQ_2011

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