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When 79.75 g of $C u S O_{4}$ sample containing impurity is reacted with KI, the liberated $I_{2}$ is reacted with 50 mL (1M) $N a_{2} S_{2} O_{3}$ in basic medium, where it oxidized into $S O_{4}^{- 2}$ ions, and $I_{2}$ reduces into $I^{-}$ , then what will be the % purity of $C u S O_{4}$ in sample?

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95%

answer is B.

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Detailed Solution

m eq. of $C u S O_{4}$ reacted = m eq. of $N a_{2} S_{2} O_{3}$ reacted $(n - f = 8) \frac{w}{159.5} = \frac{50 \times 1 \times 8}{1000}$

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