0 . 01 MKMnO _ 4 solution was added to 20 . 0mL of 0 . 05MMohr 's...

${Fe}^{2 +} + {MnO}_{4}^{-} ⟶ {Fe}^{3 +} + {Mn}^{2 +}$

$vf = 1vf = 5$

vf=Valency factor

Milli equivalents of mohr's salt = milli equivalents of ${KMnO}_{4}$

$1 \times \left[0 . 05 \times 20\right] = 5\left[0 . 01 \times {V}_{ml}\right]$

V_ml =20ml

Volume of ${KMnO}_{4}$ left = $50 - 20 = 30ml$