If valence factor (n-factor) of compound NaHC2O4⋅2H2C2O4⋅3K2C2O4⋅4Al2C2O43⋅3FeC2O4in acid base titration is x and redox titration with KMnO4 is y then value of y/x is:

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If valence factor (n-factor) of compound ${NaHC}_{2} O_{4} \cdot 2 H_{2} C_{2} O_{4} \cdot 3 K_{2} C_{2} O_{4} \cdot 4 {Al}_{2} {(C_{2} O_{4})}_{3} \cdot 3 {FeC}_{2} O_{4}$
in acid base titration is x and redox titration with ${KMnO}_{4}$ is y then value of y/x is:

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8.4

11.25

answer is B.

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Here, the value of n factor in acid base titration, $x = 5$

Here, the value of n factor in acid base titration, $y = 45$

Thus, $y / x = 45 / 5 = 9$ .

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