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Q.

Concentrated Nitric acid is act as oxidizing agent. If oxidizes $I_{2}$ to $H I O_{3},$ the n-factor of the $I_{2}$ is x, and it oxidizes $S_{8}$ to $H_{2} S O_{4},$ the n-factor of $S_{8}$ is y. What is the value of $x + y .$

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a

48

b

38

c

26

d

58

answer is C.

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Detailed Solution

$\begin{array}{l} {I_{- 2}}^{(0)} \to H I_{-}^{(+ 5)} O_{3} \\ E W o f I_{2} = \frac{M W}{n - f a c t o r} = \frac{M W}{10} \\ \Rightarrow n f a c t o r o f I_{2} = 10 = x \\ {S_{8}}_{-}^{(o)} \to H_{2} \underline{S} O_{4} \\ E W o f S_{8} = \frac{M W}{48} \\ \Rightarrow n - f a c t o r o f S_{8} = 48 = y \\ ∴ x + y = 10 + 48 \\ = 58 \end{array}$

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