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Question

The equivalent weight of $$MnS{O}_{4}$$ is half its molecular weight when it is converted to:


A
Mn2O3
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B
MnO2
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C
MnO4
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D
MnO24
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Solution

The correct option is B $$Mn{O}_{2}$$
$$\overset { +2 }{ Mn } S{ O }_{ 4 }^{ 2- }\longrightarrow \overset { +4 }{ Mn } { O }_{ 2 }+2{ e }^{ - }$$
Here, change in oxidation of $$Mn =2$$.

So, $$n-$$factor $$=2$$.

Equivalent weight $$= \dfrac{ M }{ 2 }$$

Therefore, the equivalent weight of $${ MnSO }_{ 4 }$$ is half of its molecular weight when it is converted to $${ MnO }_{ 2 }$$

Chemistry

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