Alum is found to contain hydrated monovalent cation ${[M {(H_{2} O)}_{6}]}^{+}$ , trivalent cation ${[M {(H_{2} O)}_{6}]}^{3 +}$ and ${S O}_{4}^{2 -}$ in the ratio of:

1 : 1 : 1

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1 : 2 : 3

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1 : 3 : 2

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1 : 1 : 2

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Solution

The correct option is D 1 : 1 : 2
Solution:- (D) $1 : 1 : 2$
The formula of alum is $K_{2} S O_{4} . {A l}_{2} {(S O_{4})}_{3} .24 H_{2} O$ .
There are $2$ monovalent cation, $2$ trivalent cations and $4 {S O_{4}}^{- 2}$ ion.
Thus the ratio of hydrated monovalent cation, trivalent ion and ${S O_{4}}^{- 2}$ ion is $1 : 1 : 2$ .

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