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3-92-g-of-ferrous-ammonium-sulphate-are-dissolved-in-100-ml-of-water-20-ml-of-this-solution-requires-18-ml-of-potassium-permanganate-during-titration-for-complete-oxidation-The-weight-of-KMnO-4-pre




Question Number 18186 by Tinkutara last updated on 16/Jul/17
3.92 g of ferrous ammonium sulphate  are dissolved in 100 ml of water. 20 ml  of this solution requires 18 ml of  potassium permanganate during  titration for complete oxidation. The  weight of KMnO_4  present in one litre  of the solution is
3.92gofferrousammoniumsulphatearedissolvedin100mlofwater.20mlofthissolutionrequires18mlofpotassiumpermanganateduringtitrationforcompleteoxidation.TheweightofKMnO4presentinonelitreofthesolutionis
Answered by sandy_suhendra last updated on 16/Jul/17
Commented by Tinkutara last updated on 16/Jul/17
Thanks Sir!
ThanksSir!
Commented by sandy_suhendra last updated on 16/Jul/17
3.92 gram ferrous amonium sulfate = ((3.92)/(392)) = 0.01 mole  in 20 ml water = ((20)/(100))×0.01 = 0.002 mole  mole KMnO_4  = (2/(10))×0.002 = 4×10^(−4)  mole  KMnO_4  in 1 litre = ((1000)/(18))×4.10^(−4)   = 0.022 mole/litre                         = 0.022 × (39+55+4.16) = 3.476 gram/litre
3.92gramferrousamoniumsulfate=3.92392=0.01molein20mlwater=20100×0.01=0.002molemoleKMnO4=210×0.002=4×104moleKMnO4in1litre=100018×4.104=0.022mole/litre=0.022×(39+55+4.16)=3.476gram/litre
Answered by Tinkutara last updated on 20/Jul/17

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