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One gram of $N{a_3}As{O_4}$ is boiled with excess of solid $KI$ in the presence of strong $HCl$ . The iodine evolved is absorbed in the $KI$ solution and titrated against the $0.2N$ hypo solution. Assuming the reaction to be:
$AsO_4^{3 - } + 2{H^ + } + 2{I^ - } \to AsO_3^{3 - } + {H_2}O + {I_2}$
The volume of thiosulphate hypo consumed is $(As = 75)$ :
A. $48.1ml$
B. $38.4ml$
C. $24.7ml$
D. $30.3ml$

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Last updated date: 20th Jun 2024
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Answer
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Hint: This question is based upon redox reactions of physical chemistry. Before attempting this question one must have prior knowledge of normality, n-factor and all the other terms used in this topic.

Complete step by step answer:
We will solve this question by the method of n-factor.
N factor considered as a conversion factor which is calculated by dividing molar mass of substance to get equivalent mass of the substance. N factor depends on the nature of substance which changes from one condition to another condition.
The balanced chemical equation is:
$N{a_3}AsO_4^{} + 2HCl + 2KI \to N{a_3}AsO_3^{} + 2KCl + {H_2}O + {I_2}$
N factor is calculated by the difference in oxidation number;
N factor of $N{a_3}As{O_4} = 5 - 3 \Rightarrow 2$
Equivalent weight of $N{a_3}As{O_4}$ is calculated using the formula-
${\text{Equivalent weight = }}\dfrac{{{\text{atomic mass}}}}{{n - factor}}$
Equivalent weight of $N{a_3}As{O_4} = \dfrac{{208}}{2} \Rightarrow 104$
Equivalent weight is defined as the mass of one equivalent that is the mass of a given substance which will either combine with or displace a fixed quantity of another substance. Equivalent weight is also known as gram equivalent.
Normality is defined as the ratio of mass of a substance and equivalent weight.
Generally, normality is explained as the number of gram or mole equivalents of solute present in one litre of a solution. When we say equivalent, it is the number of moles of reactive units in a compound.
Normality of $N{a_3}As{O_4} = \dfrac{1}{{104}}$
Equation of normality:
$\
  {N_1}{V_1}(Thiosulphate) = {N_2}{V_2}(N{a_3}As{O_4}) \\
   \Rightarrow 0.2 \times {V_1} = \dfrac{1}{{104}} \times 1000 \\
   \Rightarrow {V_1} = 48.1ml \\
\ $
So, the correct answer is Option A.

Note: Normality is defined as the number of equivalent weights (or simply equivalents) of solute dissolved per liter of solution. It is the gram equivalent weight of a solute per liter of solution.