An accurate weight of 0.96 g steel alloy was dissolved in concentrated nitric ac

Question

An accurate weight of 0.96 g steel alloy was dissolved in concentrated nitric acid, treated with an excess of potassium periodate, and heated to oxidize manganese present to permanganate ions according to the following reaction: 5IO_4 + Mn^2+ + 3H_2O 2MnO_4^- + 5IO_3^- + 6H^+ The resulting solution was diluted to 50 mL in a calibrated volumetric flask. Color match was achieved between the diluted solution and a 4.5 Times 10^-4 mol L^-1 KMnO_4 solution when the light path through the standard was 5.0 cm and that through the sample solution was 2. Calculate the manganese content in terms of percentage and ppm units.

Explanation / Answer

Given the concentraton of the known KMnO4 solution = 4.5x10-4 mol/L

Hence the concentration of the 50 mL of the unknown KMnO4 solution = 4.5x10-4 mol/L x (2/5) = 1.8x10-4 mol/L

Hence moles of KMnO4 in 50 mL of the solution = 1.8x10-4 mol/L x 0.050 L = 9.0x10-6 mol KMnO4

Since 1 mol of KMnO4 contains 1 mol of Mn - atom,

moles of Mn-atom in the steel alloy = 9.0x10-6 mol Mn

Atomic mass of Mn = 54.94 g/mol

Hence mass of Mn - atom in the alloy = 9.0x10-6 mol x 54.94 g/mol = 4.9446x10-4 g Mn

total mass of alloy = 0.96 g

Hence mass percentage of Mn in the alloy = (4.9446x10-4 g / 0.96 g) x 100 = 0.0515 % (answer)

Mn content in terms of ppm = (4.9446x10-4 g / 0.96 g) x 106 ppm = 515 ppm (answer)

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