3. A 0.1 M-KMnO4 solution is used for the following titration. What volume of the solution will be required to react with 0.158 g of Na2S₂O3? S₂O3 +MnO4+H₂O →MnO₂ (s) + SO42- + OH- (A) 80 ml (C) 13.33 ml (B) 26.67 ml (D) 16 ml
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Answer:
S
2
O
3
2−
→2SO
4
2−
Change in oxidation number of sulphur per molecule of S
2
O
3
2−
=2×(6−2)=8
Change in oxidation number of Mn per molecule of MnO
4
−
=7−4=3
No. of moles in 0.158 g of Na
2
S
2
O
3
=
158
0.158
=1×10
−3
No. of equivalents =8×10
−3
Normality of 0.1 M KMnO
4
solution =0.1×3=0.3
Let V mL of volume of KMnO
4
be required; then
1000
V
×0.3=8×10
−3
or V=
0.3
8
×10
−3
=10
3
=26.7 mL.