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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

The correct Answer is:
4.48
`2Kl + H_(2)SO_(4)+ H_(2)O_(2) to K_(2)SO_(4)+ 2H_(2)O + I_(2)`
From the above equation
`H_(2)O_(2) -=I_(2)` or 34g of `I_(2)`
`therefore 0.508 g` of `I_(2)` will be liberated from `H_(2)O_(2)`
`=(34)/(254)xx0.508`
`=0.068` g
The decomposition of `H_(2)O_(2)` occurs as
`underset(2xx34=68 g )(2H_(2)O_(2))to underset(22400 cm^(3) " at " NTP)(2H_(2)O) + O_(2)`
`therefore 0.068 of `H_(2)O_(2)` upon decomposition will give `O_(2)=(22400)/(68)xx0.068`
`=22.4 cm^(3)`
Now `5.0cm^(3)` of `H_(2)O_(2)` solution gives `O_(2)`
`=22.4 cm^(3)` at STP
`therefore 1.0 cm^(3) ` of `H_(2)O_(2)` solution will give `O_(2)`
`=(22.4)/(5)=4.48 cm^(8)` at STP
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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
    • PRADEEP-HYDROGEN -COMPETITION FOCUS (Numerical Value Type Questions)
    1. A 5.0 mL of solution of H(2)O(2) liberates 0.508 g of iodine from acid...

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