How much oxygen witt be liberated at $N . T . P$ , by $2 A$ current passing for $3$ minutes and $13$ seconds through acidulated water? (Equivalent weight of oxygen = $8$ )

0.032 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.541 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.629 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.713 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.032 g$
From Faraday's law of Electrolysis,

$m = \frac{c u r r e n t \times t i m e \times E q . w t .}{F}$

$\Rightarrow m = \frac{2 \times 193 \times 8}{96500}$

$\Rightarrow m = 0.032 g$

Suggest Corrections

Similar questions

O_{2}

5

193

d m^{3}

1 a m p

160

50

Acidulated water is electrolysed by 1A current for 16 minutes and 5 seconds using inert electrodes. The volume of gases liberated at STP will be :

Join BYJU'S Learning Program