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How many milliliters of 0.125 M KMnO_4 are required to react completely with 25.0 mL of 0.250 MFeSO_4 solution in the acidic medium ? |
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Answer» Solution :The balanced chemical EQUATION is : `MnO_4^(-) +5Fe^(2+) +8H^(+) rarr MN^(2+) +5Fe^(3+) +4H_2O` From the balanced equation it is clear `{:(1 mol KMnO_4 =, 5 mol FeSO_4),(158,5xx152):}` Moles of `FeSO_4` present in 25 mL of 0.250 M solution : `=(0.250)/1000xx25=0.00625` mol Let us determine the number of moles of `KMnO_4` that must react 5 mol of `FeSO_4` react with `KMnO_4` = 1mol 0.00625 mol of `FeSO_4` will react with `KMnO_4` `=1/5xx0.00625=0.00125 mol` Now, we are the calculate the volume of 1.25 M `KMnO_4` solution which contain 0.00125 mol . According to definition of morality , 0.125 M solution means that `0.125 mol KMnO_4` is present in = 1000 mL 0.00125 mol of `KMnO_4` is present in `=1000/(0.125)xx0.00125=10 mL` This can also be CALCULATED as : Morality `=("Moles"xx1000)/(Volume)` `0.125=(0.00125xx1000)/(volume)` `:. Volume = (0.00125 xx1000)/(0.125) = 10 ` mL |
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