Correct option is (b) 50
MnO2 + 4HCI → 2MnCl2 + Cl2 + 2H2O;
2KI + Cl2 → 2KCl + I2
Moles of 12 liberated moles of Cl2 = moles of MnO2 = \(\frac{6.35}{254}\)
Weight of pure MnO2 = \(\frac{6.35 \times 87}{254}\) = 2.175
% Purity of MnO2 = \(\frac{2.175 \times 100}{4.35} \) = 50