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Question: The equivalent weight of \[Zn{\left( {OH} \right)_2}\]​ in the following reaction is equal to its : ...

The equivalent weight of Zn(OH)2Zn{\left( {OH} \right)_2}​ in the following reaction is equal to its :
[Zn(OH)2+HNO3Zn(OH)(NO3)+H2O][Zn{(OH)_2} + HN{O_3} \to Zn(OH)(N{O_3}) + {H_2}O]
A.formula  wt.2\dfrac{{formula\;wt.}}{2}
B.formula  wt.1\dfrac{{formula\;wt.}}{1}
C.3×formula  wt.3 \times formula\;wt.
D.2×formula  wt.2 \times formula\;wt.

Explanation

Solution

Equivalent weight (which is otherwise called gram comparable) is denoted by E and is defined by the mass of one identical that is the mass of a given substance which combines with or displaces a fixed amount of some other substance. The equivalent weight of a component is the mass which consolidates with or dislodges 1.008 gram1.008{\text{ }}gram of hydrogen or 35.5 grams35.5{\text{ }}grams of chlorine or 8.0 grams8.0{\text{ }}grams of oxygen.

Complete step by step answer:
For a redox equivalent reaction, where a similar compound goes through decrease and oxidation all the while, the nfactorn - factor assumes a critical job in deciding the equivalent weight of the compound going through redox response.
The equivalent weight of a compound can be determined by separating the sub-atomic mass by the quantity of positive or negative electrical charges that result from the disintegration of the compound.
The numerical formula of Eq. weightEq.{\text{ }}weight is given by:
Equivalent weight = E=MwnEquivalent{\text{ }}weight{\text{ }} = {\text{ }}E = \dfrac{{{M_w}}}{n}
Where, Mw={M_w} = Molecular weight
n=n = nfactorn - factor or valency or the quantity of electrons moved.
Equivalent weight, E= Formula weightnfactor.E = {\text{ }}\dfrac{{Formula{\text{ }}weight}}{{n - factor}}.
The reaction of Zinc hydroxide with nitric acid produces zinc hydroxide nitrate with the release of water molecules. Now we have the reaction as follows:
Zn(OH)2+HNO3Zn(OH)(NO3)+H2O.  Zn{(OH)_2} + HN{O_3} \to Zn(OH)(N{O_3}) + {H_2}O.\;
Here the Zn(OH)2Zn{\left( {OH} \right)_2}discharges solo OHO{H^ - }ion in the reaction.
so nn - factor is11.
So, equivalent weight of Zn{\left( {OH} \right)_2}$$$$ = \dfrac{{Formula{\text{ }}weight}}{{n - factor}} = \dfrac{{Formula{\text{ }}weight}}{1}.
Therefore, the correct option is B. formula  wt1\dfrac{{formula\;wt}}{1}.​

Note:
The practice of equivalent weights in overall chemistry has mostly been outdated by the practice of molar masses. Equivalent weights might be estimated from molar masses if the chemistry of the matter is well identified.
In acidbaseacid - base reactions, the equivalent weight of an acid // base is the mass of which provisions or responds through a single mole of hydrogen cations (H+{H^ + }). For redox reactions, the equivalent weight of individually reactant provisions or reacts with a single mole of electrons (e{e^ - }) in a redox reaction.