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If 25.8 ml of 0.101 M $K_{2} {Cr}_{2} O_{7}$ is required to titrate 10.0 ml of a liquid iron supplement, calculate the concentration of iron in vitamin solution

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0.780 M

0.261 M

$4.35 \times 10^{- 4} M$

1.56 M

answer is D.

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Detailed Solution

${\overset{+ 3}{Cr}}_{2} O_{7}^{- 2} + {Fe}^{+ 2} \to 2 {Cr}^{+ 3} + {Fe}^{+ 3}$

Equivalent of $K_{2} {Cr}_{2} O_{7}$ = Eq. of Fe

$\Rightarrow 25.8 \times 0.101 \times 6 = 10.0 \times M \times 1 \Rightarrow M = 1.56 mol / L$

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