JEE Advance - Chemistry (2021 - Paper 2 Online - No. 11)
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 ______.
(Assume : KMnO4 reacts only with Fe2+ in the solution
Use : Molar mass of iron as 56 g mol$$-$$1)
The value of x is ______.
Answer
1.875
1.88
- OR
Explanation
$$\mathop {Fe}\limits_{(x \times {{10}^{ - 2}}mol)} + 2HCl \to \mathop {FeC{l_2}}\limits_{(x \times {{10}^{ - 2}}mol)} + {H_2}$$
Concentration of Fe2+, $${M_1} = {{x \times {{10}^{ - 2}}} \over {250}} \times 1000$$
$$ = {{10x} \over {250}}M = {x \over {25}}M$$
Volume of Fe2+ solution titrated, V1 = 25.0 mL
Concentration of $$MnO_4^ - $$, M2 = 0.03 M
Volume of $$MnO_4^ - $$ used, V2 = 12.5 mL
$$5F{e^{2 + }} + MnO_4^ - \to M{n^{2 + }} + 5F{e^{3 + }}$$
Using relation,
$${{{M_1}{V_1}} \over {{n_1}}} = {{{M_2}{V_2}} \over {{n_2}}}$$
n2M1V1 = n1M2V2 [n1 and n2 are number of moles of Fe2+ and MnO$$_4^ - $$ reacting]
$$1 \times {x \over {25}} \times 25 = 5 \times 0.03 \times 12.5 \Rightarrow x = 1.88$$
The volume of x is 1.88.
Concentration of Fe2+, $${M_1} = {{x \times {{10}^{ - 2}}} \over {250}} \times 1000$$
$$ = {{10x} \over {250}}M = {x \over {25}}M$$
Volume of Fe2+ solution titrated, V1 = 25.0 mL
Concentration of $$MnO_4^ - $$, M2 = 0.03 M
Volume of $$MnO_4^ - $$ used, V2 = 12.5 mL
$$5F{e^{2 + }} + MnO_4^ - \to M{n^{2 + }} + 5F{e^{3 + }}$$
Using relation,
$${{{M_1}{V_1}} \over {{n_1}}} = {{{M_2}{V_2}} \over {{n_2}}}$$
n2M1V1 = n1M2V2 [n1 and n2 are number of moles of Fe2+ and MnO$$_4^ - $$ reacting]
$$1 \times {x \over {25}} \times 25 = 5 \times 0.03 \times 12.5 \Rightarrow x = 1.88$$
The volume of x is 1.88.
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