Question

Given,\[2.480g\]of \[KCl{O_3}\] are dissolved in conc.\[HCl\] and the solution was boiled. Chlorine gas evolved in the reactions was then passed through a solution of \[KI\]and liberated iodine was titrated with \[100\] ml of hypo. \[12.3ml\] of same hypo solution required \[24.6ml\] of \[0.5\] iodine for complete neutralization. Calculate \[\% \] purity of \[KCl{O_3}\]sample.

Answer 1

Hint: In chemistry, periodic tables play a vital role. In the periodic table there are totally \[118\] elements. In the periodic table there are totally \[18\] columns and \[7\] rows. The columns are called groups. Hence, \[18\] groups in the periodic table. The rows are called periods. Hence, totally \[7\] period in the table.

Complete answer:
The given statement is
\[12.3ml\] of same hypo solution required \[24.6ml\] of \[0.5\] iodine for complete neutralization.
\[2.480g\] of \[KCl{O_3}\] are dissolved in conc.\[HCl\] and the solution is boiled.
The chemical reaction for the above discussion is given below,
\[KCl{O_3} + HCl \to KCL + {H_2}O + C{l_2}\]
The balanced chemical equation for the above reaction is given below,
\[2KCl{O_3} + 12HCl \to 2KCL + 6{H_2}O + 6C{l_2}\]
Chlorine gas evolved in the reactions was then passed through a solution of \[KI\]and liberated iodine was titrated with \[100\] ml of hypo.
The chemical reaction for the above discussion is given below,
\[C{l_2} + KI \to KCl + {I_2}\]
The balanced chemical equation for the above reaction is given below,
\[C{l_2} + 2KI \to 2KCl + {I_2}\]
The molar equivalent of iodine released in the chemical reaction is equal to the molar equivalent of the hypo solution.
The molar equivalent of hypo = the molar equivalent of I2
The molar equivalent of hypo is \[100\].
The number of iodine atoms is \[2\].
We calculate the molar equivalent of ${I_2}$
\[ = \dfrac{{100}}{2}\]
\[ = 50\]
The molar equivalent of $C{l_2}$ = the molar equivalent of ${I_2}$
The molar equivalent of $C{l_2}$ is \[50\].
The molar equivalent of potassium hypochlorite is
\[ = \dfrac{{50 \times 2}}{6}\]
On simplification we get,
\[ = \dfrac{{50}}{3}\]
We calculate the purity of potassium hypochlorite is,
The molecular weight of potassium hypochlorite is \[122.5\].
\[\dfrac{W}{{122.5}} \times 1000 = \dfrac{{50}}{3}\]
\[W = \dfrac{{50}}{3} \times \dfrac{{122.5}}{{1000}}\]
On simplification we get,
\[W = 2.042\]
The molecular weight of potassium hypochlorite is \[2.042\].
We calculated \[\% \] purity of \[KCl{O_3}\] sample is
\[ = \dfrac{{2.042}}{{2.480}} \times 100 = 82.32\]
The percentage of purity of \[KCl{O_3}\] sample is \[82.32\% \].
According to the above discussion, we conclude the percentage of purity of \[KCl{O_3}\] sample is \[82.32\% \].

Note:
Arrhenius is one of the concepts for acid and bases. Swedish chemist Arrhenius is the world's first chemist to talk about acid and bases. Depending on his concept and limitations later so many concepts are developed. Even Arrhenius is the birthplace of the concept of acids and bases. Followed by Arrhenius chemist the concept of acids and bases discuss Bronsted and Lowry in the world.
According to the Arrhenius concept one substance said to acid means, acid is nothing but a substance that dissociates to give hydrogen ion when decomposed in the water. Examples of Arrhenius acids are hydrochloric acid (\[HCl\]), sulphuric acid (\[{H_2}S{O_4}\]), nitric acid (\[HN{O_3}\]) etc,.