Question

100mL of 0.1N hypo decolorized iodine by the addition of x g of crystalline copper sulphate to excess of KI. The value of x is:
a.) 5.0g
b.) 1.25g
c.) 2.5g
d.) 4g

Answer 1

Hint: To answer this question, we must focus on the concept of normality and how to calculate it. Normality can be described as the number of gram or mole equivalents of solute present in one liter of a solution.

Complete step by step solution:
Similar to molarity, where the molecular weight is used for calculating the molar concentration, for normal solution concentrations, we generally prefer to use the equivalent weight (EW).
We first have to find the valence factor. For this let us write the chemical reaction:
\[2CuS{O_4} + 4KI \to 2{K_2}S{O_4} + 2CuI + {I_2}\]
The change in oxidation state of Cu = valence factor = 2-1 = 1.
Next we must write the reaction for iodine produced with Hypo solution:
\[{I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_6}{O_6}\]
From this reaction we see that the hypo is decolorized due to Iodine gas produced.
So we may write,
No. of equivalents of iodine = No. of equivalents of hypo solution
And since the Iodine gas is solely produced due to reaction with hydrated copper sulphate initially. We may say,
No. of equivalents of iodine = No. of equivalents of hydrated copper sulphate.
We know, Molecular weight of hydrated copper sulphate = 250g
\[No.\,of\,equivalents\,of\,CuS{O_4}.5{H_2}O = \dfrac{{Molecular\,weight}}{{n - factor}} = \dfrac{{250}}{1} = 250\]
We know,
\[\dfrac{{weight}}{{no.\,of\,equivalents}} = Normality \times Volume\]
Therefore,
\[\dfrac{x}{{250}} = 0.1 \times 0.1\]
\[x = 2.5g\]
Hence, the correct answer is Option (C) 2.5g.

Additional information:
Normality is sometimes used in place of molarity because often 1 mole of acid does not completely neutralize 1 mole of base. Hence, in order to have a one-to-one relationship between acids and bases, many chemists prefer to express the concentration of acids and bases in normality.

Note: The normal concentration of a solution or normality is always equal to or greater than the molar concentration or molarity of a solution. The normal concentration can be calculated by multiplying the molar concentration by the number of equivalents per mole of solute.