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When 214gms of potassium iodide reacts with excess of KI in presence of H⁺ then it produces , I₂ now I₂ is completely reacted with 1M Na₂S₂O₃ sol in $O H^{-}$ (basic) medium. Where it is converted into sulphate ions. Then volume (in ml) of the hypo needed to react at the end point of the reaction is

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answer is 750.

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Detailed Solution

The balanced chemical equation

$(i) {KIO}_{3} + 5 KI + 3 H_{2} O \to 3 I_{2} + 6 KO H$

$(ii) {Na}_{2} S_{2} O_{3} + 4 I_{2} + 10 NaOH \to 2 {Na}_{2} {SO}_{4} + 8 NaI + 5 H_{2} O$

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$8 gm - eq of I_{2} \to 1 gm - eq of Hypo n - factor of I_{2} = 2$

$6 gm - eq of I_{2} \to ? 6 / 8 = 3 / 4 = 0 .75 gm - eq of Hypo.$

No of gm-equivalents $= N \times V_{is} lt \Rightarrow V_{lt} = \frac{0 .75}{1 N} = 0 .75 L = 750 ml$

$[∴ 1 {MNa}_{2} S_{2} O_{3} = 1 {NNa}_{2} S_{2} O_{3}]$

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When 214gms of potassium iodide reacts with excess of KI in presence of H+ then it produces , I2 now I2 is completely reacted with 1M Na2S2O3 sol in OH- (basic) medium. Where it is converted into sulphate ions. Then volume (in ml) of the hypo needed to react at the end point of the reaction is