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How many gram of $KmO_4$ should be taken to make up 250 ml of a solution of such strength that 1ml is equivalent to 5Mg of Fe in a) 1.414g b) 0.70g c) 3.16g d) 1.58g

Asked by Ranjeetgupta26068 | 21 May, 2019, 08:07: AM

Expert Answer

Option (b) will be right when the unit is given in kiligram.

We know,

MnO subscript 4 to the power of minus space rightwards arrow space Mn to the power of 2 plus end exponent space plus 7 space space space space space space space space space space plus 2 Change space in space oxidation space number space equals space 5 Fe to the power of 2 plus end exponent space rightwards arrow space Fe to the power of 3 plus end exponent plus 2 space space space space space space space space plus 3 Change space in space oxidation space number space equals space 1 Therfore semicolon Equivalent space weight space of space KMnO subscript 4 space equals space straight M over 5 Equivalent space weight space of space Fe space equals fraction numerator space straight M over denominator 1 end fraction straight M. space equ space of space Fe space equals space 5 over 56 1 space ml space of space straight f space KMnO subscript 4 space contains space 5 over 56 space straight M space equ space of space straight f space KMnO subscript 4 250 space ml space of space straight M space equ space of space straight f space KMnO subscript 4 space will space conatain space space 5 over 56 cross times 250 space Mequ space of space KMnO subscript 4 We space know comma Mequ space of space KMnO subscript 4 space equals space fraction numerator Weight space in space mg over denominator Equivalent space weight space in space mg end fraction therefore Weight space in space mg space equals space Mequ space of space KMnO subscript 4 space cross times Equivalent space weight space in space mg Equivalent space weight space of space space KMnO subscript 4 space in space mg space equals space fraction numerator Molecular space weight space over denominator Change space in space oxidation space no. end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 158 over 5 space equals space 31.6 space gm Equivalent space weight space of space space KMnO subscript 4 space in space mg space equals space 31.6 space cross times 10 cubed space mg Weight space of space KMnO subscript 4 space in space mg space equals space 5 over 56 cross times 250 cross times space 31.6 space cross times 10 cubed space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 705.35 space cross times 10 cubed space mg space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 705 space gm Weight space of space KMnO subscript 4 space in space mg space equals 0.7 space kg

The weight of KMnO₄ required is 705 gm or 0.7 kg

Answered by Varsha | 21 May, 2019, 13:11: PM

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