In the determination of manganese in a steel sample, the following procedure was

Question

In the determination of manganese in a steel sample, the following procedure was followed: 10367 g of steel was weighed, dissolved in acid, and treated so as W oxidize all manganese to permanganate. The sample was diluted to 250 mL in a volumetric flask. 0.89 mL of 0.1104 N KMnO4 was diluted to 100 mL in a volumetric flask, and used as the standard. The standard solution had a reading of 50.0% T. At the same wavelength the steel sample had a reading of 24.3% T. Calculate the % of Mn in the steel sample.

Explanation / Answer

Given

standard solution = 50% transmittance

steel sample = 24.3% transmittance

Standard solution concentration = 0.1104 N

grams of steel taken, 10367 g

volume of diluted sample = 250 mL = 0.25 L

amount of KMnO4 in standard = 0.1104 x 0.25 x 158.03 = 4.362 g

So the amount of KMnO4 in sample of steel = 4.362 x 24.3/50 = 2.12 g

1 mole of KMnO4 has 1 mole of Mn

or,

158.03 g of KMnO4 has 55 g of Mn

55/158.03 = x/2.12

x = 0.74 = 74% Mn is present in steel sample

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