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 12.5mL of 0.03 M KMnO4 solution to reach the end point. 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 solutionUse : Molar mass of iron as 56gmol−1)The value of x is .

A sample (5.6 g) containing iron is completely dissolved in cold dilut

1.	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 12.5mL of 0.03 M KMnO4 solution to reach the end point. 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 solutionUse : Molar mass of iron as 56gmol−1)The value of x is .
Answer» A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 $m L$ of solution. Titration of 25.0 $m L$ of this solution requires $12.5 m L of 0.03 M K M n O_{4}$ solution to reach the end point. Number of moles of $F e^{2 +}$ present in 250 $m L$ solution is $x \times 10^{- 2}$ (consider complete dissolution of $F e C l_{2} .$ ) The amount of iron present in the sample is $y %$ by weight. (Assume : $K M n O_{4}$ reacts only with $F e^{2 +}$ in the solution Use : Molar mass of iron as $56 g m o l^{- 1}$ ) The value of $x$ is .