Question

What will be the volume of $O_2$ at N.T.P liberated by $5$ A current flowing for $193$ s through acidulated water?

• A. $56$ ml
• B. $112$ ml
• C. $158$ ml
• D. $965$ ml

Answer 1

The correct option is A $56$ ml

Number of moles of electrons $=\frac{5\times 193}{96500}=0.01moles$

$4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$

4 moles of electrons liberate 1 mole of oxygen. Hence, 0.01 mole of electrons will liberate 0.0025 moles of oxygen.

1 mole of oxygen at N.T.P occupies a volume of 22400 mL.

0.0025 moles of oxygen at N.T.P will occupy a volume of $0.0025\times 22400=56ml$

Hence, option A is correct.