$($ મોલર દળ $Fe=56\, g\, mol^{-1}$, $Cl=35.5\, g\, mol^{-1})$
Moles of $Fe(OH)_3$ $ = \frac{{weight\,in\,g}}{{M.\,weight\,of\,Fe{{(OH)}_3}}}$
$ = \frac{{2.14\,g}}{{107\,g/mol}} = 0.02\,mol.$
$1.0$ mole of $Fe(OH)_3$ is obtained from $= 1.0$ mole of $FeCl_3$
$0.02$ mole of $Fe(OH)_3$ is obtained from $0.02$ mole of $FeCl_3$
Molarity
$ = \frac{{No.\,of\,moles}}{{Volume\,in\,L}} = \frac{{0.02\,mole}}{{0.1\,L}} = 0.2\,M$
$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
અહી $20\, mL$ of $0.1\, M\, KMnO_4$ એ કોના બરાબર હશે