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 KMnO _ 4 solution to reach the end point. Number of moles of Fe ^ 2+ present in 250 ~mL solution is x 10^ -2 (consider co

Question

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 KMnO _ 4 solution to reach the end point. Number of moles of Fe ^ 2+ present in 250 ~mL solution is x 10^ -2 (consider complete dissolution of FeCl _ 2 ). The amount of iron present in the sample is y \% by weight. (Assume: KMnO _ 4 reacts only with Fe ^ 2+ in the solution Use: Molar mass of iron as 56 ~g ~mol ^ -1 ) The value of x is________ .

Quantrex Academy · Accepted Answer

Correct answer: A. A

Moles of  Fe 2 +  present in  250 &#160; ml  solution  = x &#215; 10 - 2 Moles of  Fe 2 +  present in  25 &#160; ml  solution  = &#160; x &#215; 10 - 3 &#160; mole At equivalence point eq. of  Fe 2 +  = eq. of  KMnO 4 Moles Fe 2 + &#215; vf Fe 2 + = Moles KMnO 4 &#215; vf KMnO 4 x &#215; 10 &#8722; 3 &#215; 1 = 0 .03 &#215; 12 .5 &#215; 10 &#8722; 3 &#215; 5 x = 1 . 875 &#160; mole Moles of  Fe 2 +  present in  250 &#160; ml  solution  = 1 . 875 &#215; 10 - 2 Mass of  Fe 2 +  present in  250 &#160; ml  solution = 1 . 875 &#215; 10 - 2 &#215; 56 &#160; gm = 1 . 05 &#160; gm %  of  Fe y = 1 .05 5 .6 &#215; 100 = 18 . 75 %

JEE Advanced Chemistry Question (2021) — Solution

Question

Options

Answer

Step-by-step solution