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In iodometric estimation, the following reactions occur
$2C{u^{2 + }} + 4{I^ - } \to C{u_2}{I_2} + {I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_4}{O_6}$ $0.12mole$ Of $CuS{O_4}$ was added to excess of KI solution and the liberated iodine required $120ml$ of hypo. The molarity of hypo solution was:
A) $2$
B) $0.20$
C) $0.1$
D) $1.0$

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Hint: We know that, iodometric titration, is a volumetric qualitative analysis method, and it is a redox titration in which the disappearance of elementary iodine indicates the top point. The iodine liberated by reaction is titrated indirectly with the analyte, while the iodine is titrated directly with the titrant.

Complete step by step answer:
First, we discuss the procedure of iodometric estimation:
All the oxidizing agents oxidize iodide ion to iodine in acidic condition. The iodine formed within the reaction can then be titrated by means of a typical hypo solution. This sort of indirect titration is given the overall name of iodometry.
Iodometric methods of study have a good applicability for the subsequent reasons:
Potassium iodide, KI, is quickly available in high purity.
A good indicator, starch, is out there to signal the equivalence point within the reaction between iodine and thiosulfate. Starch turns blue-black within the presence of iodine when the blue-black color disappears then the iodine is completely reduced to the iodide ion.
Iodometric reactions are rapid and quantitative.
An exact and stable reducer, hypo \[\left( {N{a_2}{S_2}{O_3}} \right)\] is out there to react with the iodine.
The amount of iodine liberated within the reaction between iodide ion and an oxidant may be a measure of the number of oxidants originally present within the solution. The quantity of ordinary hypo solution required to titrate the liberated iodine is then like the quantity of oxidant. Iodometric methods are often used for the quantitative determination of strong oxidizing agents like salt, permanganate, peroxide, cupric ion and oxygen.
The reactions in iodometric titration is,
$2C{u^{ + 2}} + 4{I^ - } \to C{u_2}{I_2} + {I_2}$
${I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_4}{O_6}$
The number of moles pure copper sulfate is $0.12moles$.
The volume of $N{a_2}{S_2}{O_3}$ is$120ml$.
We can calculate the molarity of the solution using the formula,
${\text{Molarity}} = \dfrac{{{\text{Mass of solute}}\left( {{\text{in moles}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$
We can calculate the molarity of the hypo solution as follows,
$\Rightarrow {\text{Molarity}} = \dfrac{{0.12moles}}{{0.120L}} = 1M$
The molarity of the hypo solution is $1M$.

So, the correct answer is Option D .

We can calculate the molality using the formula,
The mathematical expression of molality is,
${\text{Molarity}}\left( {\text{m}} \right) = \dfrac{{{\text{Moles of solute}}\left( {mol} \right)}}{{{\text{Kilograms of solvent}}\left( {kg} \right)}}$.

Last updated date: 23rd Sep 2023
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