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

Step-by-step solution

Moles of Fe 2 + present in 250   ml solution = x × 10 - 2 Moles of Fe 2 + present in 25   ml solution =   x × 10 - 3   mole At equivalence point eq. of Fe 2 + = eq. of KMnO 4 Moles Fe 2 + × vf Fe 2 + = Moles KMnO 4 × vf KMnO 4 x × 10 − 3 × 1 = 0 .03 × 12 .5 × 10 − 3 × 5 x = 1 . 875   mole Moles of Fe 2 + present in 250   ml solution = 1 . 875 × 10 - 2 Mass of Fe 2 + present in 250   ml solution = 1 . 875 × 10 - 2 &