The equivalent weight of Hypo in the reaction Na2 S2O3+Cl2+H2O→Na2SO4+2HCl+S is [M = mol. wt]

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M/2

M/3

answer is B.

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Detailed Solution

Na2 S+22O3+Cl2+H2O→Na2S+6O4+2HCl+S0Total ox.state of sulphur/mole of Na2S2O3=+4Total ox.state of sulphur on RHS=+6+0 =+6Total increase in ox.state of sulphur/mole of Na2S2O3=2∴ EW of Na2 S2O3.5H2O(Hypo)=M2Alternate method: In Na2S2O3 , ox.states of sulphur are -2 and +6 In Na2SO4 ox. state of sulphur = +6A sulphur atom with '+6' ox.state in Na2S2O3 doesn't show any change in ox.state.But the sulphur atom with '-2' ox.state in Na2S2O3 changes to zero (in sulphur)Thus Total increase in ox.state of sulphur/mole of Na2S2O3=2∴ EWof Na2 S2O3=M2

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