Finding the concentration of each ion in saturated solution of lead(II) hydroxide Question

If the solubility of $\ce{Pb(OH)2}$ is $\pu{0.155 g/L},$ then the concentration of each ion in a saturated solution of $\ce{Pb(OH)2}$ is
A. $[\ce{Pb^2+}] = \pu{0.155 g/L}$ and $[\ce{OH-}] = \pu{0.155 g/L}$
B. $[\ce{Pb^2+}] = \pu{0.052 g/L}$ and $[\ce{OH-}] = \pu{0.103 g/L}$
C. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{1.29E-3 mol/L}$
D. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{6.43E-4 mol/L}$

Answer

 C. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{1.29E-3 mol/L}$

My Approach
What I did is that I wrote the dissociation equation:
$$\ce{Pb(OH)2 &lt;=&gt; Pb^2+ + 2 OH-}$$
Then I used to mole ratio to discover that:
$$
\begin{align}
γ(\ce{Pb^2+}) &amp;= \pu{0.155 g/L}\\
γ(\ce{OH-}) &amp;= \pu{0.31 g/L}
\end{align}
$$
I then converted the mass concentration $γ$ into molar concentration $c$ since the $\pu{mol/L}$ is commonly used to describe solubility:
$$\pu{0.155 g/L} × \frac{\pu{1 mol}}{\pu{207.2 g}} = \pu{7.48E-4 mol/L}$$
However, that's not the answer. What did I do wrong?
The mistake you made is that you assumed mass concentrations of the ions from the stoichiometric relation, which is only applicable to the amount of substance, and, as a consequence, amount concentration.

You have to convert mass concentration $γ$ to the molar concentration c first:

$$c(\ce{Pb(OH)2}) = \frac{γ(\ce{Pb(OH)2})}{M(\ce{Pb(OH)2})} = \frac{\pu{0.155 g L-1}}{\pu{241.21 g mol-1}} = \pu{6.43E-4 mol L-1}$$

And, since it's a saturated solution, then a nearly complete dissociation can be assumed:

$$[\ce{Pb^2+}] = c(\ce{Pb(OH)2}) = \pu{6.43E-4 mol L-1}$$

$$[\ce{OH-}] = 2 × c(\ce{Pb(OH)2}) = 2 × \pu{6.43E-4 mol L-1} = \pu{1.29E-3 mol L-1}$$