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Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptCalculation of moleMedium2 minQB

ChemistryMediumsingle choice

In a titration, $H_{2} O_{2}$ is oxidised to $O_{2}$ by $MnO_{4}^{-} .24 mL$ of $0.1 M H_{2} O_{2}$ requires 16 mL of $0.1 M MnO_{4}^{-}$ solution. Hence, $MnO_{4}^{-}$ changes to :

Options:

Answer:

Solution:

H₂O₂ $⟶$ 2H⁺ + O₂ + 2e⁻ 0.1 M = 0.2 N

MnO₄⁻ $⟶$ Mn^x+ 0.1 M = 0.1 (7 – x) N

$V_{1} N_{1} = V_{2} N_{2}$

$0.2 \times 24 = 0.1 \times (7- x) \times 16$

$x = 4$

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Calculation of mole

⏱ 2mℹ️ Source: QB

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