A bottle of H2O2 is labelled as 10V H2O2 . 112 ml of this solution of H2O2 is titrated against 0.04M. acidified solution of KMnO4. The volume of KMnO4 in litre is

5.6 volume of H2O2=1N10 volume of H2O2=105.6N meq of H2O2= meq of Mno4- 105.6×112=0.04×5(n− factor )×VV=1000ml=1Lit

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A bottle of H₂O₂ is labelled as 10V H₂O₂ . 112 ml of this solution of H₂O₂ is titrated against 0.04M. acidified solution of KMnO₄. The volume of KMnO₄ in litre is

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Detailed Solution

$\begin{array}{l} 5 .6 volume of H_{2} O_{2} = 1 N \\ 10 volume of H_{2} O_{2} = \frac{10}{5 .6} N \\ meq of H_{2} O_{2} = {meq of Mno}_{4}^{-} \\ \frac{10}{5 .6} \times 112 = 0 .04 \times 5 (n - factor) \times V \\ V = 1000 ml = 1 Lit \end{array}$

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