3.92 g of ferrous ammonium suphate ar dissolved in 100 Ml water 20 ml of this solution requires 18 complete oxidation the weight of KMnO_(4) present in one litre of the solution is

3.92 g of ferrous ammonium suphate ar dissolved in 100 Ml water 20 ml

1.	3.92 g of ferrous ammonium suphate ar dissolved in 100 Ml water 20 ml of this solution requires 18 complete oxidation the weight of KMnO_(4) present in one litre of the solution is
Answer» <html><body><p>34.76 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a><br/>12.38 g<br/>1.23 g<br/>3.476 g</p>Solution :The redox reaction involving the oxidation of `Fe^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>+)` is<br/> `MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarr5Fe^(3+)+<a href="https://interviewquestions.tuteehub.com/tag/mn-548487" style="font-weight:bold;" target="_blank" title="Click to know more about MN">MN</a>^(2+)+4H_(2)O` <br/> Mol <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a> of ferrous ammonium suphate <br/> `(NH_(4))SO_(4)FeSO_(4)6H_(23)O=392`<br/> applyinhg molarity equation <br/> `=(M_(2)V_(2))/(n_(2))` <br/> or `(0.1xx20)/(5)=(M_(2)xx18)/(1)`<br/> or `M_(2)=(0.1xx20)/(5xx18)=(M)/(45)` <br/>Amount of `KMnO_(4)` present in one litre<br/> =Molarity x mol wt <br/> `1/45xx158=3.51` g <br/> since 3.476 g isclose to 3.51 g therefore option (d) is correct</body></html>