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0.0333 M `KMnO_4` is used for the following titration.What volume of the solution in ml will be required to read with 0.158 g of `Na_2S_2O_3`? (Mol. Wt. of `Na_2S_2O_3=158`) `S_2O_3^(2-)+MnO_4^(-)+H_2O to MnO_2(s)+SO_4^(2-)+OH^(-)` (not balanced) |
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Answer» Correct Answer - 80 `S_2O_3^(2-)to2SO_4^(2-)` Change in oxidation number of sulphur per molecule of `S_2O_3^(2-)=2xx(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_2S_2O_3=0.158/158=1xx10^(-3)` No of equivalents =`8xx10^(-3)` Normality of 0.0333 M `KMnO_4` solution =`0.0333xx3=0.1` Let V ml of volume of `KMnO_4` be required, then `V/1000xx0.1=8xx10^(-3)` or V=80 ml |
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