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50 mL of an aqueous solution of H(2)O(2)...

50 mL of an aqueous solution of `H_(2)O_(2)` was treated with an excess of KI solution and dilute `H_(2)SO_(4)`. The liberated iodine required 20 mL 0.1 N `Na_(2)S_(2)O_(3)` solution for complete interaction. Calculate the concentration of `H_(2)O_(2)` in `g//L`.

Text 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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When 100 mL of an aqueous of H_(2)O_(2) is titrated with an excess of KI solution in dilute H_(2)O_(2) , the liberated I_(2) required 50 mL of 0.1 M Na_(2) S_(2)O_(3) solution for complete reaction. Calculate the percentage strength and volume strength of H_(2)O_(2) solution.

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