$(1)\;2.5g\qquad(b)\;1.5g\qquad(c)\;1.338g\qquad(d)\;1g$

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$MnSO_4.4H_2O\qquad\underrightarrow{\Delta}\qquad Mn_3O_4$+dissolved in 100ml of N/10 $FeSO_4$

Meq of $FeSO_4$ taken=$\large\frac{1}{10}$$\times 100=10$

The solution further requires 50ml of $KMnO_4$,an oxidant (of unknown N),hence $FeSO_4$ is in excess

Excess of Meq of $FeSO_4$=Meq of $KMnO_4$(50ml)

25ml of N-$KMnO_4$=30ml of N/10 $FeSO_4$

$25\times N=30\times \large\frac{1}{10}$$\Rightarrow N=\large\frac{3}{25}$

Excess meq of $FeSO_4=\large\frac{3}{25}$$\times 50=6$

Meq of $FeSO_4$ used against $Mn_3O_4$=10-6=4

Also,meq of $Mn_3O_4$=Meq of $MnSO_4$

$\large\frac{g}{E_{MnSO_4}}$$\times 1000=4$

$\Rightarrow g=\large\frac{4}{1000}$$\times E_{MnSO_4}$

$E_{MnSO_4}.H_2o=\large\frac{223}{x}$

$3Mn^{2+}+4H_2O\rightarrow Mn_3O_4+8H6++2e^-$

$x=\large\frac{2}{3}\Rightarrow E=\large\frac{223}{\Large\frac{2}{3}}$

$\Rightarrow g=\large\frac{4}{1000}\times \frac{223}{\Large\frac{2}{3}}$

$\Rightarrow 1.338gm$

Hence (c) is the correct answer.

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