To a solution of excess of KI in dilute ...

To a solution of excess of KI in dilute `H_(2)SO_(4)`. 25 mL of an unknown solution of `KMnO_(4)` were added. The liberated iodine was exactly reduced by 42.5 mL of `N//10 Na_(2)S_(2)O_(3)` solution. Calculate the concentration of `KMnO_(4)` solution.

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The correct Answer is:

`5.372 g//L`

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To a 25 mL H_2 O_2 solution, excess of acidified solution of KI was added. The iodine liberated required 20.0 mL of 0.3 N Na_2 S_2 O_3 solution. Calculate the volume strength of H_2 O_2 solution.

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 .

To a 25 mL of H_(2)O_(2) solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3 N Na_(2)S_(2)O_(3) solution. Calculate the volume strength of H_(2)O_2 solution. Strategy : Volume strength of H_(2)O_(2) solution is related to its normality by the following relation Volume strength (V)=5.6xx"Normality" (N) where, Normality =((meq)H_(2)O_(2))/V_(mL) According to the law of equivalence (meq)_(Na_(2)S_(2)O_(3))=(meq)_(I_(2))=(meq)_(H_(2)O_(2))

A sample of MnSO_(4). 4H_(2)O is strongly heated in air. The residue ( Mn_(3)O_(4)) left was dissolved in 100 mL of 0.1NFeSO_94) containing dil. H_(2)SO_(4) . This solution was completely reacted with 50 mL of KMnO_(4) solution. 25 mL of this KMnO_(4) solution was completely reduced by 30 mL of 0.1N FeSO_(4) solution. Calculate the amount of MnSO_(4).4H_(2)O in sample.

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