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Q.

How many grams of I₂ are present in a solution which requires 40 mL of 0.11 N Na₂S₂O₃ to react with it? (l = 727)
$S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}$

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a

12.7 g

b

0.558 g

c

25.4 g

d

11.4 g

answer is B.

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Detailed Solution

Eq. of l₂ = Eq. of $S_{2} O_{3}^{2 -}$
$\frac{w}{254} \times 2 = \frac{40}{1000} \times 0 .11 \Rightarrow w = 0 .558 g$

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