Question

0.1 M $KMnO_{4}$ is used for the following titration. What volume of the solution in ml will be required to react with 0.158 g $Na_{2}S_{2}O_{3}.5H_{2}O$ ?
$3S_{2}O_{3}^{2 -}+MnO_{4}^{-}+H_{2}O \rightarrow 8MnO_{2}+6SO_{4}^{2 -}+2OH^{-}$

• A. 1.7 ml
• B. 0.17 ml
• C. 17 ml
• D. 1.07 ml

Answer 1

Answer: (C)

As per as the balanced equation given below
$3S_{2}O_{3}^{2 -}+8MnO_{4}^{-}+H_{2}O \rightarrow 8MnO_{2}+6SO_{4}^{2 -}+2OH^{-}$
$3$ moles of $Na_{2}S_{2}O_{3}=8$ moles of $KMnO_{4}$
$3\times 248$ g of $Na_{2}S_{2}O_{3}=8\times 158$ g of $KMnO_{4}$
0.158 g of $Na_{2}S_{2}O_{3}=\frac{8 \times 158 \times 0.158}{3 \times 248}$
$=0.27$ g $KMnO_{4}$
Volume $=\frac{W t . \times 1000}{M o l a r i t y \times M o l . W t .}$
$=\frac{0.27 \times 1000}{0.1 \times 158}=17 ml$ .