Solubility product constant $$\left( {{K_{sp}}} \right)$$  of salts of types $$MX,M{X_2}\,{\text{and}}\,{M_3}X$$   at temperature T are 4.0 × 10^-8, 3.2 × 10^-14 & 2.7 × 10^-15, respectively. Solubilities ( mol dm^-3 ) of the salts at temperature $$'T'$$ are in the order —

Solution :
$$MX \to \mathop {{M^ + }}\limits_s + \mathop {{X^ - }}\limits_s $$     $${\text{(Where sis the solubility)}}$$
$$\eqalign{ & {\text{Then}}\,{K_{sp}} = {S^2}\,\,\,\,{\text{or}}\,s = \sqrt {{K_{sp}}} \cr & {\text{Similarly for}}\,M{X_2} \to {M^{2 + }} + 2{X^ - } \cr & {K_{sp}} = s \times {\left( {2s} \right)^2} = 4{s^3}\,\,\,{\text{or}}\,\,s = {\left[ {\frac{{{K_{sp}}}}{4}} \right]^{\frac{1}{3}}} \cr} $$
\[\text{and}\,\text{for}\,\,{{M}_{3}}X\to 3{{\underset{3s}{\mathop{M}}\,}^{+}}+{{\underset{s}{\mathop{X}}\,}^{-3}}\]
$${K_{sp}} = {\left( {3s} \right)^3} \times s = 27{s^4}\,\,\,s = {\left[ {\frac{{{K_{sp}}}}{{27}}} \right]^{\frac{1}{4}}}$$
From the given values of $${{K_{sp}}}$$  for $$MX,M{X_2}\,{\text{and}}\,{M_3}X$$     we can find the solubilities of those salts at temperature, $$T.$$
$$\eqalign{ & {\text{Solubility}}\,{\text{of}}\,MX{\text{ = }}\sqrt {4 \times {{10}^{ - 8}}} = 2 \times {10^{ - 4}} \cr & {\text{Solubility}}\,{\text{of }}M{X_2} = {\left[ {\frac{{3.2 \times {{10}^{ - 14}}}}{4}} \right]^{\frac{1}{3}}}\,\,\,{\text{or }}{\left[ {\frac{{32}}{4} \times {{10}^{ - 15}}} \right]^{\frac{1}{3}}} \cr & = {\left[ {8 \times {{10}^{ - 15}}} \right]^{\frac{1}{3}\,\,}}\,\,\,\,\,\,{\text{or}}\,\,\,\,\,2 \times {10^{ - 15}} \cr & {\text{Solubility of}}\,{M_3}X = {\left[ {\frac{{2.7 \times {{10}^{ - 15}}}}{{27}}} \right]^{\frac{1}{4}}} \cr & = {\left[ {{{10}^{ - 16}}} \right]^{\frac{1}{4}}}\,\,\,{\text{or}}\,\,\,\,{10^{ - 4}} \cr} $$
Thus the solubilities are in the order $$MX > {M_3}X > M{X_2}$$
i.e the correct t anser is (D) .