Increase in the oxidation state of each reductant is
1.
Net increase in the oxidation state of two atoms in one molecule of reductant is
2.
Net decrease in the oxidation state of
KMnO4 in acidic medium is
5.
Thus,
2 moles
KMnO4≡5 moles reductant.
Let
M g/mol be the molar mass of the reductant.
Number of moles of KMnO4=201000×0.01M=2×10−4 moles
Number of moles of reductant =52×2×10−4 moles =5×10−4 moles
Mass of reductant =5×10−4 moles ×M
But 2.52 g of reductant is present in 1L.
Mass of reductant in 25 mL =251000×2.52=0.063 g
Hence, 0.063 g =5×10−4 moles ×M
or, M=126gmol−1
Hence, the molecular weight is 126 g/mol.