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The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. `Mn_(2)O_(3)`B. `MnO_(2)`C. `MnO_(4)^(-)`D. `MnO_(4)^(-)` |
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Answer» Correct Answer - B Equivalent weight in redox system is defined as : `E = ("Molar mass")/("n-factor")` Here n-factor is the net change in oxidation number per formula unit of oxidising or reducing agent. In the present case, n-factor is because equivalent weight is half of molecular weight. Also, `{:("n-factor",MnSO_(4),rarr,(1)/(2)Mn_(2)O_(3),1(+2 rarr +3)),(,MnSO_(4),rarr,MnO_(2),2(+2 rarr +4)),(,MnSO_(4),rarr ,MnO_(4)^(-),5(+2 rarr +7)),(,MnSO_(4),rarr,MnO_(4)^(2-),4(+2 rarr +6)):}` Therefore, `MnSO_(4)`converts to `MnO_(2)`. |
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