A sample (5.6 g) containing iron is completely dissolved in cold dilute HC1 to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO 4 solution to reach the end point. Number of moles of Fe 2+ present in 250 mL solution is x x 10 -2 (consider complete dissolution of FeCl 2 ). The amount of iron present in the sample is y% by weight. (Assume: KMnO4 reacts only with Fe 2+ in the solution Use: Molar mass of iron as 56 g mol -1 )
(i) The value of x is_______.
Text Solution
Verified by Experts1.875
(1.875)
meq of Fe +2 = meq of KMnO 4
x 
10 -2 x 1000x1 = 12.5x0.03x5x10 x= 1.875 mole
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems