An aqueous solution containing $0.1 g {KIO}_{3}$ (formula weight= 214) and an excess of KI was acidified with HCl. The liberated 12 consumed 45 mL of thiosulphate. The molarity of sodium thiosulphate solution is : The reaction involved is :
${IO}_{3}^{-} + I^{-} + H^{+} ⟶ I_{2} + H_{2} O$

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0.0623 M

0.252 M

0.0313M

0.126 M

answer is A.

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Detailed Solution

The molarity of sodium thiosulphate solution can be calculated as,

$\begin{array}{l} {IO}_{3}^{-} + 5 I^{-} + 6 H^{+} ⟶ 3 I_{2} + 3 H_{2} O \\ n_{I_{2}} = 3 \times \frac{0.1}{214}; \frac{3 \times 0.1}{214} \times 2 = \frac{45 \times M \times 1}{1000} \\ \Rightarrow M = 0.0623 \end{array}$

Thus, the correct option is 0.0623 M

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