One gram of Na 3 AsO 4 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 43-+2 H ++2 I - longrightarrow AsO 33-+ H 2 O + I 2 calculate the volume of thiosulphate hypo consumed. [Atomic weight of As =75]

Question

One gram of Na 3 AsO 4 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 43-+2 H ++2 I - longrightarrow AsO 33-+ H 2 O + I 2 calculate the volume of thiosulphate hypo consumed. [Atomic weight of As =75] (A) 48.1 mL (B) 38.4 mL (C) 24.7 mL (D) 30.3 mL

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/4315b43a511e1ddee17a47d6d4e09c04-.png" /> molar mass Na 3 AsO 4=23 × 3+75+16 × 4=208 eq. of AsO 4-=(1/((208)2))=((1/104)) equivalent of Na 3 AsO 4= equivalent of I 2= equivalent of Na 2 S 2 O 3 (1/104 × .2) L=V=48.1 mL

48.1 mL

38.4 mL

24.7 mL

30.3 mL

molar mass Na3​AsO4​=23×3+75+16×4=208 eq. of AsO4−​=(2208​)1​=(1041​) equivalent of Na3​AsO4​= equivalent of I2​= equivalent of Na2​S2​O3​ 104×.21​L=V=48.1mL

molar mass $N a_{3} A s O_{4} = 23 \times 3 + 75 + 16 \times 4 = 208$

eq. of $A s O_{4}^{-} = \frac{1}{( \frac{208}{2} )} = (\frac{1}{104})$

equivalent of $N a_{3} A s O_{4} =$ equivalent of $I_{2} =$ equivalent of $N a_{2} S_{2} O_{3}$

$\frac{1}{104 \times .2} L = V = 48.1 m L$