Question
Question: In the reaction \({{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}\) , t...
In the reaction I2+2Na2S2O3→2NaI+Na2S4O6 , the equivalent weight of sodium thiosulphate is:
A. Molecular weight
B. 2Molecular weight
C. 3Molecular weight
D. 4Molecular weight
Solution
We were asked to find the equivalent weight of sodium thiosulphate. So, we first discuss equivalent weight and molecular weight briefly. Then we discuss how to calculate the equivalent weight of a compound in a chemical equation.
Complete step by step answer:
I2+2Na2S2O3→2NaI+Na2S4O6
-Equivalent weight: equivalent is that the mass of 1 equivalent that's the mass of a given substance which can combine with or displace a hard and fast quantity of another substance. The equivalent of a component is that the mass which mixes with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. These values correspond to the relative atomic mass divided by the standard valence, for oxygen as example that's 216.0g.
For acid-base reactions, the equivalent of an acid or base is that the mass which supplies or reacts with one mole of hydrogen cations. For redox reactions, the equivalent weight of each reactant supplies or reacts with one mole of electrons in a redox reaction.
-Molecular weight: The molecular mass is the mass of a given molecule. Different molecules of an equivalent compound may have different molecular masses because they contain different isotopes of a component . The related quantity relative molecular mass, as defined by IUPAC, is that the ratio of the mass of a molecule to the unified mass unit. The molecular mass and relative molecular mass are distinct from but associated with the molar mass. The molar mass is defined because the mass of a given substance divided by the quantity of a substance and is expressed in g/mol.
Now, let us consider the given chemical equation,
I2+2Na2S2O3→2NaI+Na2S4O6
Here, Change in oxidation state of sulphur is 0.5
As there are two sulphur atoms per molecule, the entire change within the oxidation number for hypo during this reaction is 1.
So, we get,