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Q.
What will be the volume of $O _{2}$ Liberated at $NTP$ by passing $5 A$ current For $193\sec$. through acidified water.
NTA AbhyasNTA Abhyas 2020Electrochemistry
Solution:
$W =\frac{ i \times t \times E }{ F }$
$=\frac{5 \times 193 \times 8}{96500}$
$\because$ At NTP, volume of $32\, g \, O _{2}=22400 \, mL$
$\therefore $ Volume of $0.08\, g\, O _{2}=\frac{22400 \times 0.08}{32}$
$=56\, mL$