One gram of Na₃AsO₄ is boiled with excess of solid Kl in presence of strong HCl. The iodine evolved is absorbed in Kl solution and titrated against 0.2 N hyposolution.
Assuming the reaction to be
${AsO}_{4}^{3 -} + 2 H^{+} + 2 I^{-} ⟶ {AsO}_{3}^{3 -} + H_{2} O + I_{2}$
calculate the volume of thiosulphate hypo consumed.
[Atomic weight of As = 75]

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48.1

38.4 mL

24.7 mL

30.3 mL

answer is A.

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Detailed Solution

$molar mass {Na}_{3} {AsO}_{4} = 23 \times 3 + 75 + 16 \times 4 = 208$

$eq. of {AsO}_{4}^{- 3} = \frac{1}{(\frac{208}{2})} = (\frac{1}{104})$

$\begin{array}{l} equivalent of {Na}_{3} {AsO}_{4} = equivalent of I_{2} = {equivalent of Na}_{2} S_{2} O_{3} \end{array}$

$\frac{1}{104 \times .2} L = \frac{1000}{104 x 0.2} ml = 48.0769 ml$

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