Single Choice

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
Correct Answer
B$$112$$ ml
C$$158$$ ml
D$$965$$ ml

Solution

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

$$\displaystyle 4OH^- \rightarrow 2H_2O + O_2 + 4e^-$$

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 $$\displaystyle 0.0025 \times 22400 = 56 \: ml$$

Hence, option A is correct.

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