JEE Advance - Chemistry (2021 - Paper 2 Online - No. 12)
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 KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x $$\times$$ 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 solution
Use : Molar mass of iron as 56 g mol$$-$$1)
The value of y is ______.
(Assume : KMnO4 reacts only with Fe2+ in the solution
Use : Molar mass of iron as 56 g mol$$-$$1)
The value of y is ______.
Answer
18.75
Explanation
Mass of sample = 5.6 g
Number of moles of Fe = 1.88 $$\times$$ 10$$-$$2 mol
Molar mass of Fe = 56 g/mol
Mass of Fe = Moles of Fe $$\times$$ Molar mass
= 1.88 $$\times$$ 10$$-$$2 $$\times$$ 56 = 1.05 g
% Fe = $${{Mass\,of\,Fe} \over {Mass\,of\,sample}} \times 100 = {{1.05} \over {5.6}} \times 100$$
$$\Rightarrow$$ y = 18.75%
The value of y is 18.75.
Number of moles of Fe = 1.88 $$\times$$ 10$$-$$2 mol
Molar mass of Fe = 56 g/mol
Mass of Fe = Moles of Fe $$\times$$ Molar mass
= 1.88 $$\times$$ 10$$-$$2 $$\times$$ 56 = 1.05 g
% Fe = $${{Mass\,of\,Fe} \over {Mass\,of\,sample}} \times 100 = {{1.05} \over {5.6}} \times 100$$
$$\Rightarrow$$ y = 18.75%
The value of y is 18.75.
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