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# A sample of $MnSO_44H_2O$ is heated in air to give $Mn_3O_4$.The residue $Mn_3O_4$ is dissolved in 100ml of $N/10$ $FeSO_4$ containing $H_2SO_4$.The solution reacts completely with 50ml of $KMnO_4$.25ml of this $KMnO_4$ requires 30ml of N/10 $FeSO_4$ for complete oxidation.Determine the amount of $MnSO_4.4H_2O$ is the sample

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

$MnSO_4.4H_2O\qquad\underrightarrow{\Delta}\qquad Mn_3O_4$+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}$
$E_{MnSO_4}.H_2o=\large\frac{223}{x}$
$3Mn^{2+}+4H_2O\rightarrow Mn_3O_4+8H6++2e^-$
$x=\large\frac{2}{3}\Rightarrow E=\large\frac{223}{\Large\frac{2}{3}}$
$\Rightarrow g=\large\frac{4}{1000}\times \frac{223}{\Large\frac{2}{3}}$
$\Rightarrow 1.338gm$