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 .

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 . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

meq of Fe +2= meq of KMnO 4 x × 10-2 × 1000 × 1=12.5 × 0.03 × 5 × 10 x =1.875 mole

meq of Fe+2= meq of KMnO4​ x×10−2×1000×1=12.5×0.03×5×10 x=1.875mole

meq of $F e^{+ 2} =$ meq of $K M n O_{4}$
$x \times 1 0^{- 2} \times 1000 \times 1 = 12.5 \times 0.03 \times 5 \times 10$
$x = 1.875 m o l e$