
What is the equivalent weight of $ {\text{KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2} $ as an oxidant in pressure of 4.0 N HCl when ICl becomes the reduced form? $ ({\text{K}} = 39.0,{\text{I}} = 127.0). $
Answer
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Hint: The mass of a molecule is measured in moles (m). Since they contain various isotopes of an atom, different molecules of the same compound can have different molecular masses. The related quantity relative molecular mass is a unitless ratio of a molecule's mass to the unified atomic mass unit, as described by IUPAC.
Complete answer:
The mass of one counterpart, that is, the mass of a given substance that can blend with or displace a fixed quantity of another substance, is known as equivalent weight. The density of an element that mixes with or displaces 1.008 grams of hydrogen, 8.0 grams of oxygen, or 35.5 grams of chlorine is its equivalent weight.
We know that
$ {\text{Equivalent mass }} = \dfrac{{{\text{ Molecular mass }}}}{{{\text{ change in O}}{\text{.N}}{\text{. per molecule }}}} $
The degree of oxidation (loss of electrons) of an atom in a chemical compound is defined by the oxidation state, also known as the oxidation number. The oxidation state, which may be positive, negative, or zero, is the potential charge that an atom would have if all of its bonds to other atoms were entirely ionic, with no covalent component. When it comes to genuine bonds, this is never the case.
Now in $ {\text{KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2} $
In $ {\text{KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2},{\text{I}}{{\text{O}}_3}{\text{ }} $ is present as $ {\text{I}}{{\text{O}}_3}^ - $ i.e,
The oxidation state of $I$ is $ x - 6 = - 1{\text{ or }}x = + 5 $
$ {\text{KH}}{\left( {{{\text{I}}^{ + 1}}{{\text{O}}_3}} \right)_2} \to 2{{\text{I}}^{ + 1}}{\text{Cl}} $
Decrease in oxidation state is calculated as $= +10 – 2 = 8$.
$ {\text{ Equivalent wt}}{\text{. of KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2} = \dfrac{{{\text{ Molecular weight}}}}{8} $
$ \therefore {\text{ Equivalent wt}}{\text{. of KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2} = \dfrac{{39 + 12(127 + 48)}}{8} = 48.75 $ .
Note:
The molecular mass and relative molecular mass are not the same as the molar mass, but they are related. The molar mass is expressed in g/mol and is defined as the mass of a given substance divided by the volume of that substance. For interacting with macroscopic (weighable) volumes of a substance, the molar mass is typically the more accurate figure.
Complete answer:
The mass of one counterpart, that is, the mass of a given substance that can blend with or displace a fixed quantity of another substance, is known as equivalent weight. The density of an element that mixes with or displaces 1.008 grams of hydrogen, 8.0 grams of oxygen, or 35.5 grams of chlorine is its equivalent weight.
We know that
$ {\text{Equivalent mass }} = \dfrac{{{\text{ Molecular mass }}}}{{{\text{ change in O}}{\text{.N}}{\text{. per molecule }}}} $
The degree of oxidation (loss of electrons) of an atom in a chemical compound is defined by the oxidation state, also known as the oxidation number. The oxidation state, which may be positive, negative, or zero, is the potential charge that an atom would have if all of its bonds to other atoms were entirely ionic, with no covalent component. When it comes to genuine bonds, this is never the case.
Now in $ {\text{KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2} $
In $ {\text{KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2},{\text{I}}{{\text{O}}_3}{\text{ }} $ is present as $ {\text{I}}{{\text{O}}_3}^ - $ i.e,
The oxidation state of $I$ is $ x - 6 = - 1{\text{ or }}x = + 5 $
$ {\text{KH}}{\left( {{{\text{I}}^{ + 1}}{{\text{O}}_3}} \right)_2} \to 2{{\text{I}}^{ + 1}}{\text{Cl}} $
Decrease in oxidation state is calculated as $= +10 – 2 = 8$.
$ {\text{ Equivalent wt}}{\text{. of KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2} = \dfrac{{{\text{ Molecular weight}}}}{8} $
$ \therefore {\text{ Equivalent wt}}{\text{. of KH}}{\left( {{\text{I}}{{\text{O}}_3}} \right)_2} = \dfrac{{39 + 12(127 + 48)}}{8} = 48.75 $ .
Note:
The molecular mass and relative molecular mass are not the same as the molar mass, but they are related. The molar mass is expressed in g/mol and is defined as the mass of a given substance divided by the volume of that substance. For interacting with macroscopic (weighable) volumes of a substance, the molar mass is typically the more accurate figure.
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