$(a)\;4\qquad(b)\;5\qquad(c)\;3\qquad(d)\;6$

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$Fe_2O_3+Zn\rightarrow Fe^{2+}\rightarrow 100ml$ solution

25ml sample=17ml of 0.0167M of an oxidant.

Let 'n' be the no of electrons taken up by oxidant.

Now Meq of $Fe^{2+}$ in25ml=Meq of oxidant

$\Rightarrow [0.0167\times n]\times 17$

Meq of $Fe^{2+}$ in 100ml=$[0.0167\times n\times 17]\times 4$

Also meq of $Fe_2O_3$=Meq of $Fe^{2+}$ in 100ml

$\Rightarrow 0.0167\times 68]n$

Meq of $Fe_2O_3=\large\frac{0.552}{E}$$\times 1000$

$\Rightarrow (0.0167\times 68)n$

$E_{\large Fe_2O_3}=?$

1 mole of $Fe_2O_3=2Fe^{3+};Fe^{3+}+1e^-\rightarrow Fe^{2+}$

$X=2$ for 1 mole of $Fe_2O_3$

$\Rightarrow E_{Fe_2O_3}=\large\frac{160}{2}$$=80$

$n=\big[\large\frac{0.552}{80}$$\times 1000\big]/[0.0167\times 68]=6$

$\Rightarrow$ Electrons taken by Oxidant=6

Hence (d) is the correct answer.

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