Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

$2.56 \times 10^{- 3}$ equivalent of $KOH$ is required to neutralise $0 .12544 g H_{2} {XO}_{4}$ . The atomic mass of $X$ (in $g/mol$ ) is:
[Given: $H_{2} {XO}_{4}$ is a dibasic acid]

Options:

Answer:

Solution:

Let the atomic weight of $X = y$
No. of equivalents of $KOH$ = No. of equivalents of $H_{2} {XO}_{4}$
i.e. $\frac{0 .12544}{\frac{66 + y}{2}} = 0 .12544$
$∴ y = 32$

Stream:NTA_ABHYASSubject:ChemistryTopic:Redox ReactionsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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