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(D) For $PQ$ (type $AB$): $K_{sp} = s^2$. So,$s = \sqrt{K_{sp}} = \sqrt{4.0 	imes 10^{-20}} = 2.0 	imes 10^{-10} \ M$.For $PQ_2$ (type $AB_2$): $K_{

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$2 > 3 > 1$

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(D) For $PQ$ (type $AB$): $K_{sp} = s^2$. So,$s = \sqrt{K_{sp}} = \sqrt{4.0 	imes 10^{-20}} = 2.0 	imes 10^{-10} \ M$.For $PQ_2$ (type $AB_2$): $K_{sp} = 4s^3$. So,$s = (K_{sp}/4)^{1/3} = (3.2 	imes 10^{-14} / 4)^{1/3} = (8.0 	imes 10^{-15})^{1/3} = 2.0 	imes 10^{-5} \ M$.For $PQ_3$ (type $AB_3$): $K_{sp} = 27s^4$. So,$s = (K_{sp}/27)^{1/4} = (2.7 	imes 10^{-35} / 27)^{1/4} = (1.0 	imes 10^{-36})^{1/4} = 1.0 	imes 10^{-9} \ M$.Comparing the molar solubilities: $2.0 	imes 10^{-5} (2) > 1.0 	imes 10^{-9} (3) > 2.0 	imes 10^{-10} (1)$.Thus,the decreasing order is $2 > 3 > 1$.

$S.No.$	$Formula$	$K_{sp}$
$1$	$PQ$	$4.0 \times 10^{-20}$
$2$	$PQ_2$	$3.2 \times 10^{-14}$
$3$	$PQ_3$	$2.7 \times 10^{-35}$

The solubility products of three sparingly soluble salts are given below. What is the correct decreasing order of their molar solubility?
$S.No.$ $Formula$ $K_{sp}$
$1$ $PQ$ $4.0 \times 10^{-20}$
$2$ $PQ_2$ $3.2 \times 10^{-14}$
$3$ $PQ_3$ $2.7 \times 10^{-35}$

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The solubility products of three sparingly soluble salts are given below. What is the correct decreasing order of their molar solubility?$S.No.$$Formula$$K_{sp}$$1$$PQ$$4.0 \times 10^{-20}$$2$$PQ_2$$3.2 \times 10^{-14}$$3$$PQ_3$$2.7 \times 10^{-35}$