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Q.

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

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a

M

b

c

d

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

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