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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

$$\mathrm{3}.\mathrm{92}\:\mathrm{g}\:\mathrm{of}\:\mathrm{ferrous}\:\mathrm{ammonium}\:\mathrm{sulphate} \\ $$$$\mathrm{are}\:\mathrm{dissolved}\:\mathrm{in}\:\mathrm{100}\:\mathrm{ml}\:\mathrm{of}\:\mathrm{water}.\:\mathrm{20}\:\mathrm{ml} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{solution}\:\mathrm{requires}\:\mathrm{18}\:\mathrm{ml}\:\mathrm{of} \\ $$$$\mathrm{potassium}\:\mathrm{permanganate}\:\mathrm{during} \\ $$$$\mathrm{titration}\:\mathrm{for}\:\mathrm{complete}\:\mathrm{oxidation}.\:\mathrm{The} \\ $$$$\mathrm{weight}\:\mathrm{of}\:\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{present}\:\mathrm{in}\:\mathrm{one}\:\mathrm{litre} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is} \\ $$

Answered by sandy_suhendra last updated on 16/Jul/17

Commented by Tinkutara last updated on 16/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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

$$\mathrm{3}.\mathrm{92}\:\mathrm{gram}\:\mathrm{ferrous}\:\mathrm{amonium}\:\mathrm{sulfate}\:=\:\frac{\mathrm{3}.\mathrm{92}}{\mathrm{392}}\:=\:\mathrm{0}.\mathrm{01}\:\mathrm{mole} \\ $$$$\mathrm{in}\:\mathrm{20}\:\mathrm{ml}\:\mathrm{water}\:=\:\frac{\mathrm{20}}{\mathrm{100}}×\mathrm{0}.\mathrm{01}\:=\:\mathrm{0}.\mathrm{002}\:\mathrm{mole} \\ $$$$\mathrm{mole}\:\mathrm{KMnO}_{\mathrm{4}} \:=\:\frac{\mathrm{2}}{\mathrm{10}}×\mathrm{0}.\mathrm{002}\:=\:\mathrm{4}×\mathrm{10}^{−\mathrm{4}} \:\mathrm{mole} \\ $$$$\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{in}\:\mathrm{1}\:\mathrm{litre}\:=\:\frac{\mathrm{1000}}{\mathrm{18}}×\mathrm{4}.\mathrm{10}^{−\mathrm{4}} \:\:=\:\mathrm{0}.\mathrm{022}\:\mathrm{mole}/\mathrm{litre}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{0}.\mathrm{022}\:×\:\left(\mathrm{39}+\mathrm{55}+\mathrm{4}.\mathrm{16}\right)\:=\:\mathrm{3}.\mathrm{476}\:\mathrm{gram}/\mathrm{litre}\:\:\:\:\: \\ $$$$ \\ $$

Answered by Tinkutara last updated on 20/Jul/17

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