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

Question

What will be the volume of O 2 Liberated at NTP by passing 5 A current For 193 sec. through acidified water. (A) 56 mL (B) 112 mL (C) 158 mL (D) 965 mL

Tardigrade Admin · Accepted Answer

W =( i × t × E / F ) =(5 × 193 × 8/96500) ∵ At NTP, volume of 32 g O 2=22400 mL ∴ Volume of 0.08 g O 2=(22400 × 0.08/32) =56 mL

Q. What will be the volume of $O_{2}$ Liberated at $NTP$ by passing $5 A$ current For $193 sec$ . through acidified water.

56 mL

112 mL

158 mL

965 mL

$W = \frac{i \times t \times E}{F}$
$= \frac{5 \times 193 \times 8}{96500}$
$∵$ At NTP, volume of $32 g O_{2} = 22400 m L$
$∴$ Volume of $0.08 g O_{2} = \frac{22400 \times 0.08}{32}$
$= 56 m L$

Q. What will be the volume of O2​ Liberated at NTP by passing 5A current For 193sec. through acidified water.