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A 5.0 mL of solution of H(2)O(2) liberat...

A `5.0 mL` of solution of `H_(2)O_(2)` liberates `0.508 g` of iodine from acidified `KI` solution. Calculate the strength of `H_(2)O_(2)` solution in terms of volume strength at `STP`.

Text Solution

Verified by Experts

`mEq of I_(2)=0.508/127xx1000`
`mEq of KI=0.508/127xx1000`
`mEq of H_(2)O_(2)=0.508/127xx1000`
Normality of `H_(2)O_(2)=0.503/127xx1000/5`
For `H_(2)O_(2)`
`5.6xx` Normality =Volume strength
`:.` Volume strength =`5.6xx(0.508xx1000)/(127xx5)`
`=4.48 V`
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Knowledge Check

  • 5.0 cm^(3) of H_(2)O_(2) liberates 0.508 g of iodine from an acidified KI solution. The strength of H_(2)O_(2) solution in terms of volume strenth at STP is

    A
    6.48 volumes
    B
    4.48 volumes
    C
    7.68 volumes
    D
    none of these

  • A 5.0 cm^(3) solution of H_(2)O_(2) liberates 1.27 g of iodine from an acidified KI solution. The precentage strength of H_(2)O_(2) is

    A
    `11.2`
    B
    `5.6`
    C
    `1.7`
    D
    `3.4`

  • A 5.0 mL solution of H_(2)O_(2) liberates 1.27 g of iodine from an acidified KI solution. The percentage strenth of H_(2)O_(2) is

    A
    11.2
    B
    5.6
    C
    1.7
    D
    3.4