Ions given out by Carnallite in solution are

K^{+}, M g^{2 +}, C l^{-}

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K^{+}, C l^{-}, S O_{4}^{2 -}, B r^{-}

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K^{+}, M g^{2 +} C O_{3}^{2 -}

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K^{+}, M g^{2 +}, C l^{-}, B r^{-}

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Solution

The correct option is A $K^{+}, M g^{2 +}, C l^{-}$
It is a double salt with molecular formula KCl ∙ $M g C l_{2}$ ∙ $6 H_{2} O$

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Similar questions

(N a^{+}, M g^{2 +}, C l^{-}, S O_{4}^{2 -})

i) N e i i) O^{2 -} i i i) M g^{2 +} i v) F

v) A l^{3 +} v i) C l^{-} v i i) K v i i i) N a

Species	Concentration (milliequivalent/L)
Chloride $(C l^{-})$	15
Sulphate $(S O_{4}^{2 -})$	15
Carbonate $(C O_{3}^{2 -})$	05
Bicarbonate $(H C O_{3}^{-})$	30
Calcium $(C a^{2 +})$	12
Magnesium $(M g^{2 +})$	18
$p H$	8.5

C a C O_{3}

(I) I < B r < F < C l increasing electron gain enthalpy

(I I) L i < N a < K < R b increasing metallic radius

(I I I) B < C < N < O increasing first ionization energy

(I V) A l^{3 +} < M g^{2 +} < N a^{+} < F^{-} increasing ionic size

\begin{matrix} C l^{-}, A r, K^{+}, O^{2 -}, M g^{2 +}, F^{-} \end{matrix}

C a^{2 +}

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