4.9 g of $K_{2} {Cr}_{2} O_{7}$ is taken to prepare 0.1 L of the solution 10 mL of this solution is further taken to oxidize ${Sn}^{2 +}$ ion into ${Sn}^{4 +}$ ion ${Sn}^{4 +}$ so produced is used in second reaction to prepare ${Fe}^{3 +}$ ion then the millimoles of ${Fe}^{3 +}$ ion formed will be (assume all other components are in sufficient amount) [Molar mass of $K_{2} {Cr}_{2} O_{7}$ = 294 g].

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Detailed Solution

$K_{2} {Cr}_{2} O_{7}$ + ${Sn}^{2 +}$ $\overset{}{\to}$ ${Sn}^{4 +}$ + ${Cr}^{3 +}$

${Sn}^{4 +}$ + ${Fe}^{2 +}$ $\overset{}{\to}$ ${Fe}^{3 +}$

Meq. of ${Sn}^{4 +}$ = Meq of $K_{2} {Cr}_{2} O_{7}$

Meq. of ${Sn}^{4 +}$ = Meq of ${Fe}^{3 +}$

Meq. of ${Fe}^{3 +}$ = meq of $K_{2} {Cr}_{2} O_{7}$

$N_{K_{2} {Cr}_{2} O_{7}} = \frac{4.9 \times 6}{294 \times 0.1} = 1$

Meq. of Fe⁺³ =(NV) of K₂Cr₂O₇

millimol x n-factor = 1 x 10

millimol = 10

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