Question

0.1 M $KMn{O}_4$ is used for the following titration. What volume of the solution in ml will be required to react with 0.158 g $N{a}_2{S}_2{O}_3.5H_{2}O$ ?

$3S_2{O}_3^{2-}+Mn{O}_4^{-}+H_{2}O\to 8Mn{O}_2+6S{O}_4^{2-}+2O{H}^{-}$

• 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

$3S_2{O}_3^{2-}+8Mn{O}_4^{-}+H_{2}O\to 8Mn{O}_2+6S{O}_4^{2-}+2O{H}^{-}$

$3$ moles of $N{a}_2{S}_2{O}_3=8$ moles of $KMn{O}_4$

$3\times 248$ g of $N{a}_2{S}_2{O}_3=8\times 158$ g of $KMn{O}_4$

0.158 g of $N{a}_2{S}_2{O}_3=\frac{8\times 158\times 0.158}{3\times 248}$

$=0.27$ g $KMn{O}_4$

Volume $=\frac{Wt.\times 1000}{Molarity\times Mol.Wt.}$

$=\frac{0.27\times 1000}{0.1\times 158}=17ml$ .