Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptCalculation of moleMedium2 minQB

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10 ml of $x M NaHC_{2} O_{4}$ is oxidised by 10 ml of $0.02 M MnO_{4}^{-}$ in acidic medium. Hence, value of $x$ is

Answer:

Solution:

$C_{2} O_{4}^{2 -} ⟶ C O_{2}$

$M n O_{4}^{-} ⟶ M n^{2 +}$

$10 \times x \times n - f a c t oro f O C_{2} O_{4}^{2 -} = 10 \times 0.02 \times n - f a c t oro f M n O_{4}^{-}$

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

⏱ 2mℹ️ Source: QB

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