Let the salt dissolve with molar solubility $s$.

From the given options, the possible stoichiometries are:

• $XY \rightarrow X^+ + Y^-$
• $XY_2 \rightarrow X^{2+} + 2Y^-$
• $X_2Y \rightarrow 2X^+ + Y^{2-}$

For each case, write $K_{sp}$ in terms of solubility $s$:

(i) For $XY$ $[X]=s,\quad [Y]=s$ $K_{sp}=s^2$ Unit: $M^2$

But the options for $XY$ show $M^3$, so this is inconsistent. Hence $XY$ is ruled out.

For salts of type $XY_2$ or $X_2Y$:

(ii) For $XY_2$ $XY_2 \rightleftharpoons X^{2+}+2Y^-$ $[X^{2+}] = s, \quad [Y^-]=2s$ $K_{sp}=[X^{2+}][Y^-]^2=s(2s)^2=4s^3$

(iii) For $X_2Y$ $X_2Y \rightleftharpoons 2X^+ + Y^{2-}$ $[X^+]=2s, \quad [Y^{2-}]=s$ $K_{sp}=[X^+]^2[Y^{2-}]=(2s)^2(s)=4s^3$

So both have the same relation: $K_{sp}=4s^3$

From the solubility curve (read from the graph), the molar solubility is $s = 10^{-3}\,M$

Therefore, $K_{sp}=4(10^{-3})^3=4\times 10^{-9}$

Now compare with the options:

• Option B: $XY_2,\ 1\times 10^{-9}$ → incorrect
• Option C: $XY_2,\ 4\times 10^{-9}$ → matches
• Option D: $X_2Y,\ 2\times 10^{-9}$ → incorrect value

Hence the correct option is: $\boxed{\text{C: } XY_2,\ 4\times 10^{-9}\,M^3}$

Comparison with stored correct answer:

• Stored correct answer: C
• Derived answer: C
• They agree.