Comprehension
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the endpoint. Number of moles of Fe2+ present in 250 mL solution is x × 10–2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.
(Assume: KMnO4 reacts only with Fe2+ in the solution Use: Molar mass of iron as 56 g mol–1)
(Assume: KMnO4 reacts only with Fe2+ in the solution Use: Molar mass of iron as 56 g mol–1)
Question: 1
The value of x is ________.
The value of x is ________.
Updated On: May 23, 2024
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Correct Answer: 1.875
Solution and Explanation
8H+ + 5Fe2+ + MnO – → 5Fe3+ + Mn2+ + 4H2O
For 25 ml,
meq of Fe2+ = meq of MnO –
= 12.5 × 0.03 × 5
For 250 ml,
m moles of Fe2+ = \(12.5 \times 0.03 \times 5 \times \frac{250}{25}\)
moles of Fe2+ = \(\frac{18.75}{1000}\)mol
= 18.75 × 10–3 mol
= 1.875 × 10–2 mol
x = 1.875
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Question: 2
The value of y is ________.
The value of y is ________.
Updated On: May 23, 2024
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Correct Answer: 18.75
Solution and Explanation
Weight of Fe2+ = 1.875 × 10–2 × 56 = 1.05 g
% purity of Fe2+ y = 18.75%
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