KMnO _ 4 is titrated with ferrous ammonium sulphate hexahydrate in...

By balancing redox reaction:

$10 [ {FeSO}_{4} . ( {NH}_{4}{)}_{2}{SO}_{4} . 6{H}_{2}O ] + 2{KMNO}_{4} + 8{H}_{2}{SO}_{4} \to 5{Fe}_{2} ( {SO}_{4}{)}_{3} + 2{MnSO}_{4} + 10 ( {NH}_{4}{)}_{2}{SO}_{4} + {K}_{2}{SO}_{4} + 68{H}_{2}O$

$8{H}^{+} + {MnO}_{4}^{-} + 5{Fe}^{+ 2} ⟶ 5{Fe}^{+ 3} + {Mn}^{+ 2} + 4{H}_{2}O$

For $2$ moles of $MnO{4}^{-}, 8$moles of ${H}_{2}O .$ are released $5$ moles of Mohr's salt contain $30$ moles of ${H}_{2}O .$ So, $10$ moles of Mohr's salt contain $60$ moles of ${H}_{2}O$ So, total moles of ${H}_{2}O$ produced$= 68$