What will be the volume of O 2 Liberated at NTP by passing 5 A current For 193 Sec. through acidified water.

W = i × t × E F = 5 × 1 9 3 × 8 9 6 5 0 0 ∵ At NTP, volume of 32 g O 2 = 22400 mL ∴ Volume of 0.08 g O 2 = 2 2 4 0 0 × 0 . 0 8 3 2 = 56 mL

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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

What will be the volume of O₂Liberated at NTP by passing 5 A current For 193 Sec. through acidified water.

Options:

Answer:

Solution:

$W = \frac{i × t × E}{F}$

$= \frac{5 \times 193 \times 8}{96500}$

$∵$ At NTP, volume of 32 g O₂= 22400 mL

$∴$ Volume of 0.08 g O₂ $= \frac{22400 \times 0.08}{32}$

$= 56 mL$

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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