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When 25 " mL of " an aqueous solution of...

When 25 " mL of " an aqueous solution of `H_2O_2` is titrated with an excess of KI solution in dilute `H_2SO_4`, the liberated `I_2` required 20 " mL of " 0.3 N`Na_2S_2O_3` solution for complete reaction.volume strength of `H_2O_2` solution.

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`H_(2)O_(2) + 2Kl + H_(2)SO_(4) to K_(2)SO_(4) + 2H_(2)O + l_(2)`
`2Na_(2)S_(2)O_(3) + l_(2) to Na_(2)S_(4)O_(6) + 2Nal `
Eq. mass `H_(2_O_(2) = 34/2 = 17`
20 mL `0.1 N Na_(2)S_(2)O_(3) = 20 mL 0.1 l_(2)` solution
`-= 20 mL 0.1 N H_(2)O_(2)` solution
Amount of `H_(2)O_(2) ` in mL aq. solution ` = (0.1xx17)/1000 xx 20 = 0.034 `g
Concentration in g/L = `(0.034)/50 xx 1000 = 0.68`
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