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 y 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 y is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Moles of Fe +2= x × 10-2=1.875 × 10-2 wt. of Fe +2=1.875 × 10-2 × 56 Hence percentage of Fe +2 =(1.875 × 10-2 × 56/5.6) × 100 % =18.75 %

Moles of Fe+2=x×10−2=1.875×10−2 wt. of Fe+2=1.875×10−2×56 Hence percentage of Fe+2 =5.61.875×10−2×56​×100% =18.75%

Moles of $F e^{+ 2} = x \times 1 0^{- 2} = 1.875 \times 1 0^{- 2}$
wt. of $F e^{+ 2} = 1.875 \times 1 0^{- 2} \times 56$
Hence percentage of $F e^{+ 2}$
$= \frac{1.875 \times 1 0 ^{- 2} \times 56}{5.6} \times 100%$
$= 18.75%$