Question

$0 .1 M$ $KMnO_{4}$ is used for following titration. What volume of the solution in $mL$ will be required to react with $0 .158 $ of $N a_{2}S_{2}O_{3}$ ?
Not balanced: $S_{2}O_{3}^{2 -}+MnO_{4}^{-}+H_{2}O \rightarrow MnO_{2 \left(\right. s \left.\right)}+SO_{4}^{2 -}+OH^{-}$

• A. 26.7 mL
• B. 50 mL
• C. 65 mL
• D. 75 mL

Answer 1

Answer: (A)

'$n$' factor of $N a_2 S_2 O_3=8$
'$n$' factor of $KMnO _4=3$
m.eq of $Na _2 S _2 O _3=$ m.eq of $KMnO _4$
$\frac{0.158}{158} \times 8 \times 1000=0.1 \times V \times 3$
$V =26.7\, mL$