0 .1 M KMnO4 is used for following titration. What volume of the solution in mL will be required to react with 0 .158 of N a2S2O3 ? Not balanced: S2O32 -+MnO4-+H2O arrow MnO2 (. s .)+SO42 -+OH-

Question

0 .1 M KMnO4 is used for following titration. What volume of the solution in mL will be required to react with 0 .158 of N a2S2O3 ? Not balanced: S2O32 -+MnO4-+H2O arrow MnO2 (. s .)+SO42 -+OH- (A) 26.7 mL (B) 50 mL (C) 65 mL (D) 75 mL

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/2d5b8764d345838fda8bad6a01509e11-.png" /> 'n' factor of N a2 S2 O3=8 'n' factor of KMnO 4=3 m.eq of Na 2 S 2 O 3= m.eq of KMnO 4 (0.158/158) × 8 × 1000=0.1 × V × 3 V =26.7 mL

Q. $0.1 M$ $K M n O_{4}$ is used for following titration. What volume of the solution in $m L$ will be required to react with $0.158$ of $N a_{2} S_{2} O_{3}$ ?
Not balanced: $S_{2} O_{3}^{2 -} + M n O_{4}^{-} + H_{2} O \to M n O_{2 (s)} + S O_{4}^{2 -} + O H^{-}$

26.7 mL

50 mL

65 mL

75 mL

' $n$ ' factor of $N a_{2} S_{2} O_{3} = 8$
' $n$ ' factor of $K M n O_{4} = 3$
m.eq of $N a_{2} S_{2} O_{3} =$ m.eq of $K M n O_{4}$
$\frac{0.158}{158} \times 8 \times 1000 = 0.1 \times V \times 3$
$V = 26.7 m L$

Q. 0.1M KMnO4​ is used for following titration. What volume of the solution in mL will be required to react with 0.158 of Na2​S2​O3​ ? Not balanced: S2​O32−​+MnO4−​+H2​O→MnO2(s)​+SO42−​+OH−