If the solubility of $L i_{3} N a_{3} (A I F_{6})_{2}$ is $x m o l L^{- 1}$ , then its solubility product is equal to.

12 x^{3}

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18 x^{3}

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x^{8}

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2916 x^{8}

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Solution

The correct option is D $2916 x^{8}$
When put in water, the compound $L i_{3} N a_{3} (A I F_{6})_{2}$ dissociates as:
$L i_{3} N a_{3} (A I F_{6})_{2} \to 3 L i + 3 N a + 2 (A I F_{6})$
Hence, its $K_{s} p$ can be written as
$K_{s} p$ = $[3 x]^{3} \times [3 x]^{3} \times [2 x]^{2}$
=> $K_{s} p$ = $81 x^{3} \times 81 x^{3} \times 4 x^{2}$
=> $K_{s} p$ = $2916 x^{8}$

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If the solubility of Li3Na3(AIF6)2 is xmolL−1, then its solubility product is equal to.

The correct option is D 2916x8When put in water, the compound Li3Na3(AIF6)2 dissociates as:Li3Na3(AIF6)2→3Li+3Na+2(AIF6)Hence, its Ksp can be written as Ksp = [3x]3×[3x]3×[2x]2=> Ksp = 81x3×81x3×4x2=> Ksp = 2916x8

If the solubility of $L i_{3} N a_{3} (A I F_{6})_{2}$ is $x m o l L^{- 1}$ , then its solubility product is equal to.