ExamBro
ExamBro
JEE Advanced · Chemistry · 9. Redox Reactions

A sample \((5.6 \mathrm{~g})\) containing iron is completely dissolved in cold dilute \(\mathrm{HCl}\) to prepare a \(250 \mathrm{~mL}\) of solution. Titration of \(25.0 \mathrm{~mL}\) of this solution requires \(12.5 \mathrm{~mL}\) of \(0.03 \mathrm{M} \mathrm{KMnO}_{4}\) solution to reach the end point. Number of moles of \(\mathrm{Fe}^{2+}\) present in \(250 \mathrm{~mL}\) solution is \(\mathbf{x} \times 10^{-2}\) (consider complete dissolution of \(\mathrm{FeCl}_{2}\) ). The amount of iron present in the sample is \(\mathbf{y} \%\) by weight.
(Assume: \(\mathrm{KMnO}_{4}\) reacts only with \(\mathrm{Fe}^{2+}\) in the solution Use: Molar mass of iron as \(56 \mathrm{~g} \mathrm{~mol}^{-1}\) )
The value of x is________ .

  1. A 16.45
  2. B 18.75
  3. C 15.24
  4. D 15.26
Verified Solution

Answer & Solution

Correct Answer

(B) 18.75

Step-by-step Solution

Detailed explanation

Moles of Fe2+ present in 250 ml solution =x×10-2 Moles of Fe2+ present in 25 ml solution = x×10-3 mole
From JEE Advanced
Explore more questions on app