Question

The equivalent weight of mnso4 is half its molecular weight when it is converted to

Answer 1

⭕️Equivalent mass= Molecular weight Change in oxidation no. of Mn 
                           
⭕️Oxidation state of Mn in MnSO4 = +2

⭕️Oxidation state of Mn in Mn2O3= +3

⭕So therefore we can see ️change in oxidation state of Mn = 3-2=1

⭕️Oxidation state of Mn in MnO2= +4

⭕️ So therefore we can see change in oxidation state of Mn = 4-2=2

⭕️Oxidation state of Mn in MnO4-= +7

⭕So therefore we can see ️change in oxidation state of Mn = 7-2=5

⭕️Oxidation state of Mn in MnO42-= +6

️⭕So therefore we can see ️change in oxidation state of Mn = 6-2=4

✔️✔️the correct answer will be MnO2 be MnO2

$\huge\bf{\boxed {\red{\mathfrak{thank \: you :)}}}}$

Answer 2

$Equivalent \: weight = Molecular \: mass /n-factor$

n - factor is number for a elements which denotes the change in number of oxidation state of that element .

Given:
(Equivalent weight is half of molecular weight .)

equivalent = Molecular weight/2
so, n-factor = 2 

Oxidation Number of Mn+2 change with +2 

New Oxidation Number of Mn = +4 

MnSO4 → 1/2 Mn2O3 1 (+ 2 → + 3)
MnSO4 → MnO2 2 (+ 2 → + 4) MnSO4 → MnO_4^- 5 (+ 2 → + 7) MnSO4 → MnO_4^(2-) 4 (+ 2 → + 6)
Therefore,MnSO4 converts to MnO2.

OR

state of Mn in MnSo4  is Mn+(SO4)=0

so,Mn+(-2)=0 (net charge of SO4=-2)

Mn=2

now net charge of Mn in MnO2=

Mn+(-2)2 (net charge on O=-2

Mn=4

MnSO4-----MnO2

equivalent weight =molecular weight (let M)/change in the oxidation state

equivalent weight=M/(4-2)

=M/2 

Thus equivalent weight of MnSO4 is half the molecular weight when it is converted to MnO2