The equivalent weight of $F e {S O}_{4} . {({N H}_{4})}_{2} {S O}_{4} .24 H_{2} O$ is:
(At. wt of $F e = 56, S = 32, O = 16, N = 14, H = 1$ )

58

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358

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179

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89.5

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Solution

The correct option is B $358$
Equivalent weight=molar mass/oxidation state
Molar mass of the compound=716 g
Oxidation state Fe=2 because $x + 2 \times (- 2) + 2 \times (1) = 0 x = 2$
Equivalent weight= $\frac{716}{2} = 358$

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Similar questions

[F e (C N)_{6}]^{- 3} + H_{2} O_{2} + 2 O H^{-} \to 2 [F e (C N)_{6}]^{4 -} + {2 H}_{2} O + O_{2}

[G i v e n : F e = 56, C = 12, N = 14, O = 16, H = 1]

(H = 1, S = 32, O = 16)

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(B a = 137, S = 32, F e = 56, H = 1, O = 16)

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H = 1, O = 16, S = 32, F e = 56

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