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A mixture of 0.02 mole of ${KBrO}_{3}$ and 0.01 mole of $KBr$ was treated with excess of KI and acidified. The volume of $0.01 M {Na}_{2} S_{2} O_{3}$ solution required to consume the liberated iodine will be:

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800 mL

1200 mL

1000 mL

1500 mL

answer is B.

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Detailed Solution

Here, the reaction can be given as,

$\begin{array}{l} {BrO}_{3}^{-} + 5 {Br}^{-} ⟶ 3 {Br}_{2} \\ 0.02 0.01 3 \times \frac{0.01}{5} \\ {Br}_{2} + 2 KI ⟶ 2 KBr + I_{2} \\ \frac{3 \times 0.01}{5} \times 2 = \frac{V \times 0.01 \times 1}{1000} \\ \Rightarrow V = 1200 mL \end{array}$

Thus, the required volume is 1200 mL.

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