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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 FeSo_4

  • Option 1)

    1.414g

  • Option 2)

    0.70g

  • Option 3)

    3.16g

  • Option 4)

    1.58g

 

Answers (1)

best_answer

As we lerant in 

Oxidation -

It is defined as the addition of oxygen / electronegative element to a substance or removal of hydrogen / electropositive element from a substance.

- wherein

e.g. 2H_{2}S(g) + O_{2}(g)\rightarrow 2S(s)+2H_{2}O(l)

 

 Reactions undergo as follows 

\overset{+7}{Mn}O_{4}^{-}\rightarrow \overset{+2}{}Mn^{2+}

\therefore eq. weight\ of\ KMnO_{4}=\frac{M}{5}

\overset{2+}{}Fe^{2+}\rightarrow \overset{3+}{}Fe^{3+}

Required m.eq. of Fe= 5/56 as eq. weight of Fe=56

1 ml KMnO_{4}  should contain 5/56 m equiv. of KMnO_{4}

250 ml should contain = \frac{5}{56}\times 250

Weight of  KMnO_{4}=\frac{5}{56}\times 250\times 31.6 \left ( as \frac{158}{5}=31.6 \right )=705\, \, mg=0.705g

\therefore Solution\, \, is \, \, 2


Option 1)

1.414g

Incorrect

Option 2)

0.70g

Correct

Option 3)

3.16g

Incorrect

Option 4)

1.58g

Incorrect

Posted by

Plabita

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