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H(2)O(2)+2KIoverset(40% "yield")rarrI(2)...

`H_(2)O_(2)+2KIoverset(40% "yield")rarrI_(2)+2KOH`
`H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)overset(50% "yield")rarr K_(2)SO_(4)+2MnSO_(4)+3O_(2)+4H_(2)O`
150mL of `H_(2)O_(2)` sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of `M//2H_(2)SO_(4)` for neutralisation. Other part was treated with `KMnO_(4)` yielding 6.74 litre of `O_(2)` at 1 atm. and 273 K. Using % yield indicated find volume strength of `H_(2)O_(2)` sample used.

Text Solution

Verified by Experts

The correct Answer is:
33.6
Moles of `H_(2)SO_(4)=0.1`, mole of KOH =0.2 moles of `H_(2)O_(2)` used in first reaction
`=0.2/2 =1/0.4 = 0.25`
moles of `O_(2)` produces `=6.74/22.4 = 0.3`
molarity of `H_(2)O_(2) = 0.45/0.15 = 3M` Volume strength `=11.2 xx 3 = 33.6` volumes
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