The equivalent weight of hypo in the reaction [M = molecular weight] 2Na2S2O3 + I2 → 2NaI + Na2S4O6 is

Increase in Ox.No/atom of sulphur = 0.5Total increase in Ox.No/molecule of Na2S2O3 = 1EW of Na2S2O3 = Alternate methodIncrease in Ox.No/atom of sulphur = 0.5Total increase in Ox.No/molecule of Na2S2O3 = 2EW of Na2S2O3 = = M

The equivalent weight of hypo in the reaction [M = molecular weight] 2Na2S2O3 + I2 → 2NaI + Na2S4O6 is

Increase in Ox.No/atom of sulphur = 0.5Total increase in Ox.No/molecule of Na2S2O3 = 1EW of Na2S2O3 = Alternate methodIncrease in Ox.No/atom of sulphur = 0.5Total increase in Ox.No/molecule of Na2S2O3 = 2EW of Na2S2O3 = = M

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The equivalent weight of hypo in the reaction [M = molecular weight] 2Na₂S₂O₃ + I₂ → 2NaI + Na₂S₄O₆is

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answer is A.

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Detailed Solution

$large 2N{a_2}mathop {{S_2}}limits^{ + 2} {O_3} + {text{ }}{I_2} to {text{ }}2NaI{text{ }} + {text{ }}N{a_2}mathop {{S_4}}limits^{ + 2.5} {O_6}$
Increase in Ox.No/atom of sulphur = 0.5
Total increase in Ox.No/molecule of Na₂S₂O₃ = 1
EW of Na₂S₂O₃ = $large frac{M}{1}$
Alternate method
Increase in Ox.No/atom of sulphur = 0.5
Total increase in Ox.No/molecule of Na₂S₂O₃ = 2
EW of Na₂S₂O₃ = $large frac{2M}{2}$ = M