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ppm:-      grams of a particular component present in 106 mL Solution. Here, in the original water sample Feu200bu200bu200bu200bu200b2+ concentration = 190.27 ppm The calculation is shown below, The given redon reuction-6 \mathrm{Fe}^{2+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+} \rightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}mind of \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} taken =(26 \times 0.002632) \mathrm{mmad}\left(\mathrm{rr}_{2} \mathrm{O}_{7}{ }^{2-}\right.=0.068432 \mathrm{mmol} \text { of } \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}mual of \mathrm{Fe}^{2+} reqwired for back titration =(7.89 \times 0.00886)\begin{array}{l}\therefore \text { Vnreacted en207 }{ }^{2-}=\left(\frac{0.0699054}{6}\right) \mathrm{mmol} \\=0.0116509 \mathrm{mmal} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \\\end{array}\therefore \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} reacted with \mathrm{Fe}^{2+} in wath s mple re 2+\begin{array}{l}=(0.068432-0.0116509) \mathrm{mmad} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \\=0.0567811 \mathrm{mmoN} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\end{array}\therefore 100 \mathrm{~mL} water somple conterine \mathrm{fe}^{2+}=(0.0567811 \times 6)\operatorname{mind} \mathrm{Cr}_{2} \mathrm{Of}^{2-}=0.3406866 \mathrm{mmol}\mathrm{Pe}^{2+}\begin{array}{l}\therefore 100 \mathrm{~mL} \text { water sample vontains }=\left(0.3406866 \times 10^{-3}\right) \mathrm{mod} \mathrm{Re}^{2+} \\=\left(0.3406866 \times 10^{-3} \times 55.85\right) \\g \mathrm{fe}^{2-} \\=0.019027 \mathrm{~g} F \mathrm{Fe}^{2+} \\\end{array}\therefore 100 \mathrm{~mL} wath sample contorins =0.019027 \mathrm{~g} \cdot \mathrm{Fe}^{2}+\begin{aligned}\therefore 10^{6} \mathrm{~mL} \text { water sample conton' } \mathrm{W} & =\frac{0.019027 \times 10^{6}}{100} \mathrm{~g} \mathrm{fe}^{2}+ \\& =190.27 \mathrm{~g} \mathrm{Fe}^{2+}\end{aligned}Conclusion\therefore In wate somple 190.27 \mathrm{ppm}, \mathrm{fe}^{2+} present. ...