Answer
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Hint: Molar mass is the total mass of atoms present in 1 mole of the element. Similarly, equivalent weight is the mass of one equivalent of the molecule or atom. For calculations, equivalent weight is the ratio of molecular mass and n-factor of one mole of molecules/atoms.
Complete step by step answer:
Equivalent weight also known as gram equivalent is the mass of one equivalent, that is the mass of a given substance which will combine or displace a fixed quantity of another substance in the reaction.
The equivalent weight of an element is the mass which combines with 1 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine at STP. These values correspond to the atomic weight of the molecule divided by the usual valence shown by the atom present in the molecule.
We will now write the reaction to find the equivalent weight of ${{O}_{2}}$.
${{H}_{2}}O\text{ }+\text{ }\dfrac{1}{2}{{O}_{2}}\text{ }+\text{ }2{{e}^{-}}\text{ }\to \text{ }2O{{H}^{-}}$
In the above reaction we observe that for half a mole of oxygen molecules there is a transfer of 2 electrons. So, for 1 mole of oxygen molecule the number of moles of electrons involved is 4 which is also the n-factor. The molecular mass of oxygen is 32.
Substituting the values in the formula, we get:
Equivalent weight = $\dfrac{\text{Molecular mass}}{\text{n-factor}}$
Equivalent weight = $\dfrac{32}{4}$ = 8 g
Therefore, the equivalent weight of ${{O}_{2}}$ in the reaction is 8 g.
Note: In case of acid-base reactions, the equivalent weight of an acid or base is the mass which provides or reacts with 1 mole of positive hydrogen ions.
Complete step by step answer:
Equivalent weight also known as gram equivalent is the mass of one equivalent, that is the mass of a given substance which will combine or displace a fixed quantity of another substance in the reaction.
The equivalent weight of an element is the mass which combines with 1 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine at STP. These values correspond to the atomic weight of the molecule divided by the usual valence shown by the atom present in the molecule.
We will now write the reaction to find the equivalent weight of ${{O}_{2}}$.
${{H}_{2}}O\text{ }+\text{ }\dfrac{1}{2}{{O}_{2}}\text{ }+\text{ }2{{e}^{-}}\text{ }\to \text{ }2O{{H}^{-}}$
In the above reaction we observe that for half a mole of oxygen molecules there is a transfer of 2 electrons. So, for 1 mole of oxygen molecule the number of moles of electrons involved is 4 which is also the n-factor. The molecular mass of oxygen is 32.
Substituting the values in the formula, we get:
Equivalent weight = $\dfrac{\text{Molecular mass}}{\text{n-factor}}$
Equivalent weight = $\dfrac{32}{4}$ = 8 g
Therefore, the equivalent weight of ${{O}_{2}}$ in the reaction is 8 g.
Note: In case of acid-base reactions, the equivalent weight of an acid or base is the mass which provides or reacts with 1 mole of positive hydrogen ions.
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