JEE Advanced 2021Paper 2
- Exam
- JEE Advanced
- Year
- 2021
- Paper
- Paper 2
- Subject
- Chemistry
- Chapter
- Redox Reactions
JEE Advanced 2021 · Paper 2 · Chemistry · Redox Reactions
A sample (5.6 ~g) containing iron is completely dissolved in cold...
A sample containing iron is completely dissolved in cold dilute to prepare a of solution. Titration of of this solution requires of solution to reach the end point. Number of moles of present in solution is (consider complete dissolution of ). The amount of iron present in the sample is by weight.
(Assume: reacts only with in the solution Use: Molar mass of iron as )
The value of is________ .
(Assume: reacts only with in the solution Use: Molar mass of iron as )
The value of is________ .
Correct answer
0
Step-by-step solution
Moles of present in solution
Moles of present in solution
At equivalence point
eq. of = eq. of
\left(Moles\right){Fe}^{2 +} \times \left(vf\right){Fe}^{2 +} = {\left(Moles\right)}_{{KMnO}_{4}} \times {\left(vf\right)}_{{KMnO}_{4}}x \times {10}^{- 3} \times 1 = 0 .03 \times 12 .5 \times {10}^{- 3} \times 5x = 1 . 875mole{Fe}^{2 +}250ml= 1 . 875 \times {10}^{- 2}{Fe}^{2 +}250ml= 1 . 875 \times {10}^{- 2} \times 56gm= 1 . 05gmFe\left(y\right) = \dfrac{1 .05}{5 .6} \times 100= 18 . 75 %