Question

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

A
Mn2O3
B
MnO2
C
MnO4
D
MnO24

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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