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Question

A solution contains 2.52 g of a reductant per litre. 25 mL of this solution of the reductant requires 20 mL of 0.01MKMnO4 in acidic medium for oxidation. Find the molecular weight of the reductant given that each of two atoms which undergo oxidation per molecule of the reductant suffers an increase in the oxidation state by one unit.

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Solution

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 KMnO45 moles reductant.
Let M g/mol be the molar mass of the reductant.

Number of moles of KMnO4=201000×0.01M=2×104 moles

Number of moles of reductant =52×2×104 moles =5×104 moles

Mass of reductant =5×104 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 =5×104 moles ×M

or, M=126gmol1
Hence, the molecular weight is 126 g/mol.

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