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QJEEAdv21Ch8c77: JEE Advanced 2021 Chemistry - Class 11 JEEAdv21Ch8c77 Solution: A sample (5.6 g) containing iron is completely dissolved in…

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 x is ______.

Step-by-Step Verified Solution

Fe(x×10−2mol)+2HCl→FeCl2(x×10−2mol)+H2\mathop {Fe}\limits_{(x \times {{10}^{ - 2}}mol)} + 2HCl \to \mathop {FeC{l_2}}\limits_{(x \times {{10}^{ - 2}}mol)} + {H_2}

Concentration of Fe2+, M1=x×10−2250×1000{M_1} = {{x \times {{10}^{ - 2}}} \over {250}} \times 1000

=10x250M=x25M= {{10x} \over {250}}M = {x \over {25}}M

Volume of Fe2+ solution titrated, V1 = 25.0 mL

Concentration of MnO4−MnO_4^ - , M2 = 0.03 M

Volume of MnO4−MnO_4^ - used, V2 = 12.5 mL

5Fe2++MnO4−→Mn2++5Fe3+5F{e^{2 + }} + MnO_4^ - \to M{n^{2 + }} + 5F{e^{3 + }}

Using relation,

M1V1n1=M2V2n2{{{M_1}{V_1}} \over {{n_1}}} = {{{M_2}{V_2}} \over {{n_2}}}

n2M1V1 = n1M2V2 [n1 and n2 are number of moles of Fe2+ and MnO 4−_4^ - reacting]

1×x25×25=5×0.03×12.5⇒x=1.881 \times {x \over {25}} \times 25 = 5 \times 0.03 \times 12.5 \Rightarrow x = 1.88

The volume of x is 1.88.

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